- Why the three algebraic identities of convolution — commutativity, distributivity, associativity — are not arithmetic trivia but the rules for wiring systems together.
- How a parallel connection adds impulse responses and a cascade convolves them, and why the order of blocks in an LTI cascade does not matter.
- The impulse-response tests for memory (\(h(t)=K\delta(t)\)) and causality (\(h(t)=0\) for \(t \lt 0\)), each derived from the convolution integral rather than asserted.
- The BIBO stability condition \(\int|h(\tau)|\,d\tau \lt \infty\) — and the proof that it is not merely sufficient but necessary, built by constructing the input that breaks an unstable system.
- Invertibility and deconvolution: what \(h * h_{\text{inv}} = \delta\) means, how to invert an echo, and when no inverse exists at all.
- The unit step response \(s(t)\), its derivative relation to \(h(t)\), and why laboratories measure it instead of the impulse response.
Everything Lives in the Impulse Response
Chapter 6 established that an LTI system is completely determined by its impulse response, and Chapter 7 made that concrete: the output is the convolution of the input with \(h\).
Take that claim seriously and a strong conclusion follows. Chapter 5 introduced a vocabulary for systems — memoryless or dynamic, causal or non-causal, stable or unstable, invertible or not — and defined each property in terms of the input–output relationship as a whole. But if \(h\) determines the whole relationship, then every one of those properties must be visible in \(h\) alone. There must exist a test, applied to a single signal, that answers each question. Finding those tests is half of this chapter.
The other half concerns combination. Real engineering systems are not monolithic; they are amplifiers followed by filters followed by cables, feedback paths summed with forward paths, a receiver built from a dozen blocks bought separately. If each block is LTI, the assembly is LTI, and it therefore has an impulse response of its own. What is it? The answer turns out to be written in the algebra of the convolution operator, and this is the pleasant surprise of the chapter: the identities that make convolution behave like multiplication are exactly the rules of system interconnection.
Commutativity, distributivity and associativity are usually presented as a list of properties to memorise. They are better read as three engineering statements: the roles of signal and system are interchangeable; parallel blocks add; cascaded blocks convolve. We take them in that order, prove each one, and then turn to the four structural questions.
Commutativity: Signal and System Change Places
The convolution integral treats \(x\) and \(h\) very differently on the page — one is evaluated at \(\tau\), the other is reversed and shifted — so it is not obvious that they can be exchanged. A single substitution shows that they can. Put \(\lambda = t - \tau\), so \(\tau = t - \lambda\) and \(d\tau = -d\lambda\); as \(\tau\) runs from \(-\infty\) to \(\infty\), \(\lambda\) runs from \(\infty\) to \(-\infty\), and the reversed limits absorb the minus sign.
The output is unchanged if the input and the impulse response swap roles. A system with impulse response \(h\) driven by \(x\) produces exactly the same signal as a system with impulse response \(x\) driven by \(h\).
That last sentence sounds like a curiosity, but it has an immediate practical use. When you evaluate a convolution graphically you must reverse and slide one of the two signals, and you are free to choose which. Reverse the simpler one. If \(x[n]\) is a three-sample burst and \(h[n]\) is an infinite exponential tail, flipping the three-sample burst leaves you with three products to add at each \(n\); flipping the exponential leaves you with an infinite sum that you must then recognise as a geometric series. The answer is identical; the labour is not.
Commutativity also explains a fact that surprises students meeting cascades for the first time. Because \(h_1 * h_2 = h_2 * h_1\), an amplifier followed by a filter has the same overall impulse response as the same filter followed by the same amplifier. Mathematically the order is irrelevant. Physically it is not always so — a real amplifier saturates, a real filter loads the stage before it — but those are departures from linearity and from the idealisation of non-interacting blocks, not failures of the theorem.
Distributivity and Systems in Parallel
Distributivity is inherited directly from the linearity of the integral. Convolving one input against a sum of two impulse responses is the same as convolving twice and adding.
Now read the two sides as circuits. On the left is a single system whose impulse response is \(h_1+h_2\). On the right, the same input drives two systems simultaneously and their outputs are summed — a parallel interconnection. The identity says the two are indistinguishable from outside.
Impulse responses of parallel branches add. The rule extends to any number of branches, and a branch that merely passes the signal through contributes \(\delta(t)\).
The rule runs in both directions, and the reverse direction is the more useful one. A given \(h\) that looks unpleasant can often be split into a sum of pieces you already know how to convolve. An impulse response such as \(h(t) = \delta(t) - 2e^{-3t}u(t)\) is exactly a direct path in parallel with a lowpass branch; you convolve the input with \(\delta(t)\) (which does nothing at all) and with the exponential (a standard integral) and add. Distributivity is what licenses that decomposition.
The direct path deserves a name. Convolution with \(\delta(t)\) returns the signal unchanged, so \(\delta(t)\) is the identity element of the convolution algebra, playing the role that 1 plays in ordinary multiplication. A system with \(h(t)=\delta(t)\) is a perfect wire.
Associativity and the Cascade
The third identity is the most consequential. Feed \(x\) into a system \(h_1\) and feed the result into a second system \(h_2\). The intermediate signal is \(w = x*h_1\) and the final output is \(y = w*h_2 = (x*h_1)*h_2\). Associativity says the two convolutions may be regrouped.
Exchanging the order of integration is legitimate whenever the signals are absolutely integrable, which is the case for every system we shall call stable. In the inner integral substitute \(\mu = \tau-\sigma\); it becomes \(\int h_1(\mu)h_2\big((t-\sigma)-\mu\big)d\mu\), which is precisely \((h_1*h_2)\) evaluated at \(t-\sigma\). The outer integral is then a convolution of \(x\) with that object.
Two systems in series behave as one system whose impulse response is the convolution of theirs. Combined with commutativity, this means the blocks of an LTI cascade may be reordered freely without changing the overall response.
The cascade rule is what makes system design modular. A designer who wants a particular overall response does not have to synthesise it in one piece; the response can be factored into simpler stages, each built and tested separately, and the factors reassembled in any convenient order. Every filter of order higher than two that you will meet in practice is built this way, from cascaded first- and second-order sections.
Two special cascades are worth naming now because we shall return to both. If \(h_2\) undoes what \(h_1\) does, the cascade is a perfect wire and \(h_1 * h_2 = \delta(t)\) — that is the inverse system of Section 8-8. And if a system is cascaded with a pure delay \(\delta(t-T)\), the output is simply delayed, since \(x(t)*\delta(t-T) = x(t-T)\); delay commutes with everything.
| Identity | Statement | What it means physically |
|---|---|---|
| Commutative | \(x*h = h*x\) | Input and impulse response are interchangeable; flip whichever is simpler |
| Distributive | \(x*(h_1+h_2) = x*h_1 + x*h_2\) | Parallel branches: \(h = h_1 + h_2\) |
| Associative | \((x*h_1)*h_2 = x*(h_1*h_2)\) | Cascade: \(h = h_1 * h_2\), in any order |
| Identity | \(x*\delta = x\) | \(\delta(t)\) is a perfect wire |
| Shift | \(x(t)*\delta(t-T) = x(t-T)\) | Convolving with a shifted impulse is a pure delay |
Memory, and the Systems That Have None
Chapter 5 called a system memoryless if its output at any instant depends only on the input at that same instant. Now impose that condition on the convolution integral and see what survives.
Write the output in the form that puts the integration variable on the impulse response: \(y(t) = \int h(\tau)\,x(t-\tau)\,d\tau\). Every value of \(\tau\) for which \(h(\tau)\) is non-zero contributes a sample of the input taken \(\tau\) seconds ago. For the output to depend on \(x(t)\) and nothing else, no such contribution may come from \(\tau \ne 0\). So \(h(\tau)\) must vanish for all \(\tau \ne 0\) — and the only object that is zero everywhere except the origin and still does something is a scaled impulse.
The only memoryless LTI system is a pure gain. Every other impulse response, however short, gives the system some memory of the past or some anticipation of the future.
This is a sharper statement than it first appears. It means that an LTI system with any dynamics at all — a single capacitor, one sample of delay — has memory, and that the whole of filtering, differentiation, integration, echo and resonance lies outside the memoryless class. Memorylessness in the LTI world is not a mild restriction; it collapses the system down to a multiplication by a constant.
Between the extremes sits the useful notion of memory span: the length of the interval on which \(h\) is non-zero. A moving-average filter \(h[n] = \tfrac13(\delta[n]+\delta[n-1]+\delta[n-2])\) remembers exactly three samples and then forgets completely; an \(RC\) lowpass with \(h(t) = \tfrac{1}{\tau}e^{-t/\tau}u(t)\) never forgets entirely, though after four or five time constants the memory is negligible. The first is a finite-memory system, the second an infinite-memory one, and that distinction reappears in Chapter 9 as the difference between FIR and IIR structures.
Causality: No Response Before the Cause
A causal system does not respond before it is excited. Apply that to the impulse response directly. The impulse \(\delta(t)\) arrives at \(t=0\); the response to it is \(h(t)\); so causality demands that \(h(t)\) be zero for all negative time. The argument is short because the impulse is the cleanest possible test: it happens at one instant and at no other, so anything the system produces before that instant is uncaused.
An LTI system is causal if and only if its impulse response is a causal signal. This is one of the few places where a property of a system and a property of a signal are the same words with the same meaning.
Imposing causality tightens the convolution limits, and the tightened forms are the ones you should actually use when computing. Since \(h(\tau)=0\) for \(\tau \lt 0\), the integral over \(\tau\) starts at zero. If in addition the input is switched on at the origin, so that \(x(\tau)=0\) for \(\tau \lt 0\), the upper limit closes at \(t\).
The finite upper limit is why so many textbook convolutions are tractable: a causal system driven by a causal input has only a finite overlap to integrate, and the answer is automatically zero for \(t \lt 0\).
Non-causal systems are not merely a mathematical fiction. A system that averages a signal symmetrically, \(y[n] = \tfrac13(x[n+1]+x[n]+x[n-1])\), needs one sample of the future and cannot run in real time — but if the entire signal is already stored on disk, "the future" is just data further along the file, and such filters are used constantly in image processing and offline audio work, where they are prized for having no phase distortion. Causality is a constraint imposed by real time, not by mathematics.
BIBO Stability
The stability question is the most important of the four and the only one whose test is not immediate. Chapter 5 defined bounded-input bounded-output stability: if every input satisfying \(|x(t)| \le B \lt \infty\) produces an output that is also bounded, the system is BIBO stable. The definition quantifies over all bounded inputs, which is why we cannot settle it by trying a few.
Start with the sufficiency direction, which is a chain of two inequalities. Take any input bounded by \(B\) and bound the magnitude of the convolution integral.
The first step is the triangle inequality for integrals; the second replaces \(|x|\) by its bound \(B\). So if the integral of \(|h|\) is finite, call it \(M\), then \(|y(t)| \le BM\) for every bounded input, and the system is stable. That establishes sufficiency.
Necessity requires more work, and it is worth doing because it turns the condition from a convenient test into an exact characterisation. Suppose \(\int|h(\tau)|d\tau = \infty\). We must produce a single bounded input whose output is unbounded. Choose the input that lines every contribution up in phase — the one that is \(+1\) wherever \(h\) is positive and \(-1\) wherever it is negative.
This input never exceeds 1 in magnitude, so it is perfectly bounded. Yet the output at the single instant \(t=0\) equals the very integral we assumed to be infinite. One bounded input, one infinite output value: the system is not BIBO stable. Since assuming a divergent integral forced instability, stability forces convergence, and the condition is necessary as well as sufficient.
An LTI system is BIBO stable exactly when its impulse response is absolutely integrable (continuous time) or absolutely summable (discrete time).
Two consequences deserve emphasis because they are the source of most errors. The first: the test is on \(|h|\), not on \(h\). An impulse response that decays is not automatically stable. The signal \(h(t) = \tfrac{1}{t}u(t-1)\) tends to zero as \(t\) grows, yet \(\int_1^{\infty} dt/t\) diverges logarithmically, so the system is unstable. Decay is necessary but the decay must be fast enough for the area to converge.
The second: a system on the boundary is unstable. The ideal integrator has \(h(t)=u(t)\), whose absolute integral is plainly infinite. A constant input of 1 volt — thoroughly bounded — produces a ramp that grows without limit. The same reasoning condemns the discrete accumulator \(h[n]=u[n]\). Marginal cases fail the test, and they fail it for a physically obvious reason.
| Impulse response | \(\int |h|\) or \(\sum |h|\) | Stable? |
|---|---|---|
| \(h(t) = e^{-2t}u(t)\) | \(1/2\) | Yes |
| \(h(t) = e^{2t}u(t)\) | Diverges | No |
| \(h(t) = u(t)\) (integrator) | Diverges | No |
| \(h(t) = \delta(t-5)\) (delay) | \(1\) | Yes |
| \(h[n] = a^n u[n]\) | \(1/(1-|a|)\) if \(|a| \lt 1\) | Yes iff \(|a| \lt 1\) |
| \(h[n] = \tfrac{1}{n}u[n-1]\) | Harmonic series — diverges | No |
| \(h[n] = \tfrac{1}{n^2}u[n-1]\) | \(\pi^2/6\) | Yes |
Invertibility and Deconvolution
A system is invertible if distinct inputs produce distinct outputs, so that the input can in principle be recovered from the output. For LTI systems the cascade rule turns this into a clean algebraic requirement. Recovering the input means finding a second LTI system \(h_{\text{inv}}\) which, placed after the first, returns the signal unchanged — that is, whose cascade with \(h\) is the identity element of the convolution algebra.
The inverse system is the convolutional reciprocal of \(h\). Finding it — recovering \(x\) from \(y\) when \(h\) is known — is called deconvolution.
The simplest non-trivial case is worth carrying in your head. A pure delay \(h(t)=\delta(t-T)\) is inverted by an advance, \(h_{\text{inv}}(t)=\delta(t+T)\), since \(\delta(t-T)*\delta(t+T)=\delta(t)\). Notice that the inverse of a causal system is here non-causal: undoing a delay requires seeing the future. That trade-off is generic, and it is why deconvolution in real time is difficult even when it is mathematically possible.
The discrete accumulator and first difference form the other standard pair. If \(h_1[n]=u[n]\) accumulates and \(h_2[n]=\delta[n]-\delta[n-1]\) differences, their cascade is \(u[n]-u[n-1]=\delta[n]\): each exactly undoes the other, which is Chapter 3's statement that the step and the impulse are a running-sum-and-difference pair, now read as a system fact.
Not every system has an inverse. If \(h\) destroys information, no amount of post-processing can restore it. Consider \(y[n]=x[n]-x[n-1]\) followed by nothing: two inputs differing by a constant produce the same output, so the constant is lost and the map is not one-to-one. In the frequency-domain language of Part 3 this will acquire a memorable form — a system is invertible only if its frequency response is non-zero at every frequency, since any frequency the system annihilates is gone for good. For now the algebraic version suffices: an inverse exists when the equation \(h * h_{\text{inv}} = \delta\) has a solution.
The Unit Step Response
The impulse response is the natural theoretical description of an LTI system and a hopeless experimental one. No laboratory can generate an impulse: the closest available is a very short, very tall pulse, and pushing a real amplifier or motor with such a thing drives it out of its linear region long before the measurement means anything. What can be applied cleanly is a step — close a switch, and the input goes from 0 to 1 and stays there.
Define the step response \(s(t)\) as the output when the input is \(u(t)\), with the system initially at rest. By commutativity we may put the step in the flipped position, and the convolution collapses into a running integral.
The middle step is worth pausing on: \(u(t-\tau)\) equals 1 precisely when \(\tau \le t\) and 0 otherwise, so multiplying by it simply truncates the upper limit. Differentiating both sides recovers the impulse response, and this is the relation that makes step testing practical.
Measure the step response, differentiate it, and you have the impulse response. The discrete versions replace integration by a running sum and differentiation by a first difference.
There is a second reason the step response is convenient. Many inputs are built from steps — a rectangular pulse is a difference of two of them, a staircase is a sum of several — and linearity plus time invariance mean the response to such an input is the same combination of shifted step responses. If \(x(t)=u(t)-u(t-2)\), then \(y(t)=s(t)-s(t-2)\) without any convolution being performed at all. Worked Example 5 uses this.
Finally, the step response reads off the two properties engineers care about most. Its final value \(s(\infty)=\int_{-\infty}^{\infty}h(\tau)d\tau\) is the system's DC gain, and how quickly it settles there is the system's speed. Those numbers are what a datasheet quotes.
Worked Examples
Problem. A discrete-time accumulator \(h_1[n]=u[n]\) is followed by a first-difference system \(h_2[n]=\delta[n]-\delta[n-1]\). Find the impulse response of the cascade and state what the pair of systems accomplishes.
Solution. By the cascade rule the overall impulse response is \(h_1 * h_2\). Convolving with \(\delta[n]\) leaves a signal alone and convolving with \(\delta[n-1]\) delays it by one sample, so distributivity gives
The last equality is Chapter 3's definition of the discrete impulse. The cascade has impulse response \(\delta[n]\), which is the identity: whatever goes in comes out unchanged. The two systems are inverses of each other, and this is the discrete-time image of the fact that differentiation undoes integration.
Note also what the individual blocks are like. The accumulator is unstable, since \(\sum_n |u[n]|\) diverges, while the difference system is stable. Cascading them produces a stable system, so stability of a cascade cannot be inferred from the stability of its parts — the unbounded growth of one stage was exactly cancelled by the other.
Problem. Two systems \(h_1[n]=\delta[n]+2\delta[n-1]\) and \(h_2[n]=\delta[n]-\delta[n-1]\) are connected in parallel, and their summed output drives a third system \(h_3[n]=\left(\tfrac12\right)^n u[n]\). Find the overall impulse response.
Solution. Work outwards. Distributivity collapses the parallel pair by addition:
Associativity then collapses the cascade by convolution. Since convolving with \(a\,\delta[n-m]\) scales by \(a\) and delays by \(m\),
Both terms can be put over the same exponential, using \(\left(\tfrac12\right)^{n-1} = 2\left(\tfrac12\right)^{n}\):
The two step functions agree for \(n \ge 1\) and differ only at \(n=0\), which is why the answer splits into an isolated sample plus a tail. Checking the first few values: \(h[0]=2\), \(h[1]=4(\tfrac12)=2\), \(h[2]=4(\tfrac14)=1\), \(h[3]=0.5\). The system is stable, since \(\sum|h[n]| = 2 + 4\sum_{n\ge1}(\tfrac12)^n = 2+4 = 6\), and causal, since \(h[n]=0\) for \(n \lt 0\).
Problem. Classify each system as memoryless or dynamic, and causal or non-causal: (a) \(h(t)=4\delta(t)\); (b) \(h(t)=e^{-|t|}\); (c) \(h[n]=\left(\tfrac12\right)^{n}u[n+1]\); (d) \(h(t)=\delta(t-3)\).
Solution (a). A scaled impulse and nothing else, so the system is memoryless — it is the gain \(y(t)=4x(t)\). Since \(h(t)=0\) for \(t \lt 0\), it is causal. Memoryless LTI systems are always causal, because they use nothing but the present.
Solution (b). Dynamic, since \(h\) is non-zero away from the origin. It is non-causal: \(h(-1)=e^{-1}\ne0\), so the system responds one second before the impulse arrives. This is the symmetric two-sided smoother, stable (its absolute integral is 2) but unrealisable in real time.
Solution (c). The step \(u[n+1]\) turns on at \(n=-1\), so \(h[-1]=\left(\tfrac12\right)^{-1}=2 \ne 0\). Dynamic and non-causal; the system needs one sample of the future. It is nonetheless stable, since the sum \(2 + \sum_{n\ge0}(\tfrac12)^n = 2+2 = 4\) is finite.
Solution (d). A pure three-second delay. It is dynamic — the output at time \(t\) is \(x(t-3)\), which is not the present input — and causal, since \(h(t)=0\) for \(t \lt 0\). A delay has memory; only a delay of zero would not.
Problem. Test each system for BIBO stability: (a) \(h(t)=e^{-2t}u(t)\); (b) \(h(t)=e^{-2t}u(-t)\); (c) \(h(t)=\tfrac{1}{t}u(t-1)\); (d) \(h[n]=\tfrac{1}{n^2}u[n-1]\).
Solution (a). The integrand is positive, so the absolute integral is the integral:
Stable, and its DC gain is that same number, \(\tfrac12\).
Solution (b). The same exponential, but now switched on for negative time only, where \(e^{-2t}\) grows as \(t\) becomes more negative. The integral \(\int_{-\infty}^{0}e^{-2\tau}d\tau\) diverges, so the system is unstable — and it is non-causal as well. Changing \(u(t)\) to \(u(-t)\) reversed both verdicts.
Solution (c). Here \(h(t)\to0\) as \(t\to\infty\), which tempts the wrong answer. Integrate honestly:
Unstable. The decay is too slow for the area to converge, which is the standard reminder that \(h(t)\to0\) is necessary but nowhere near sufficient.
Solution (d). The discrete counterpart, one power steeper: \(\sum_{n=1}^{\infty} 1/n^2 = \pi^2/6 \approx 1.645\), which converges. Stable. Compare it with \(h[n]=\tfrac1n u[n-1]\), the harmonic series, which diverges and is unstable. The two impulse responses look almost identical when plotted, and they land on opposite sides of the boundary.
Problem. A system's measured step response is \(s(t)=\left(1-e^{-3t}\right)u(t)\). Find \(h(t)\), the DC gain, and the response to the pulse \(x(t)=u(t)-u(t-2)\).
Solution. Differentiate, taking care with the product because \(s(t)\) contains a step:
The second term vanishes. By the sampling property of Chapter 3, \(f(t)\delta(t)=f(0)\delta(t)\), and here \(f(0)=1-e^{0}=0\). So \(h(t)=3e^{-3t}u(t)\), with no impulse at the origin — which had to be so, since the measured step response starts from zero with no jump.
The DC gain is \(s(\infty)=1\), confirmed by \(\int_0^\infty 3e^{-3\tau}d\tau = 1\).
For the pulse, do not convolve. The input is a difference of two shifted steps, so by linearity and time invariance the output is the same difference of shifted step responses:
Reading the two intervals separately: on \(0 \lt t \lt 2\) the output rises as \(1-e^{-3t}\), reaching \(1-e^{-6}\approx 0.9975\) at \(t=2\); after \(t=2\) the two expressions combine to \(\left(e^{6}-1\right)e^{-3t}\), a pure decay from that same value. The system charges while the pulse is present and discharges afterwards, exactly as an \(RC\) network would.
Problem. A channel adds a single delayed copy of its input: \(y(t)=x(t)+a\,x(t-T)\). Find its impulse response, construct the inverse system, and say when the inverse is usable.
Solution. Setting \(x(t)=\delta(t)\) reads the impulse response straight off: \(h(t)=\delta(t)+a\,\delta(t-T)\) — a direct path in parallel with an attenuated, delayed one.
We need \(h_{\text{inv}}\) with \(h * h_{\text{inv}} = \delta\). Try to cancel the echo by subtracting it, which introduces a second-order echo, which must itself be cancelled, and so on. That suggests an alternating series of delays:
Verify by convolving. Convolution with \(\delta(t)\) reproduces the series, and convolution with \(a\delta(t-T)\) reproduces it scaled by \(a\) and shifted by \(T\):
Re-index the second sum with \(m=k+1\); its terms are \(-(-a)^{m}\delta(t-mT)\) for \(m \ge 1\), which cancel the \(m \ge 1\) terms of the first sum exactly. All that survives is \(k=0\), namely \(\delta(t)\). The inverse is confirmed.
Usability is a stability question. The inverse has \(\int|h_{\text{inv}}| = \sum_{k\ge0}|a|^{k}\), which converges to \(1/(1-|a|)\) only when \(|a| \lt 1\). A weak echo can be removed; an echo as strong as the direct path cannot, because the corrections needed grow rather than die away. The inverse is causal here, incidentally, because the echo arrives after the direct path — had the delay been an advance, the inverse would have needed the future.
Chapter Summary
Convolution is commutative, distributive and associative — which is to say roles are interchangeable, parallel blocks add, and cascaded blocks convolve.
\(h_{\text{par}} = h_1+h_2\) and \(h_{\text{cas}} = h_1*h_2\), in any order. \(\delta(t)\) is the identity and \(\delta(t-T)\) a pure delay.
The only memoryless LTI system is a gain, \(h(t)=K\delta(t)\). Any other impulse response gives memory, and its width is the memory span.
\(h(t)=0\) for \(t \lt 0\). This closes the convolution limits to \(\int_0^t\), which is why causal problems are finite.
BIBO stable if and only if \(\int|h|\,dt \lt \infty\) or \(\sum|h[n]| \lt \infty\). Decay alone is not enough; the area must converge.
\(h*h_{\text{inv}}=\delta\) defines deconvolution; \(s(t)=\int_{-\infty}^{t}h\), so \(h=ds/dt\) recovers the impulse response from a measurable test.
Practice Problems
Problems 1 to 3 exercise the interconnection rules; 4 to 6 are the standard classification drills that GATE asks in one line; 7 and 8 ask you to prove things. In every classification question, name the test you are applying before you apply it.
- Three systems are connected so that \(h_1[n]\) and \(h_2[n]\) are in parallel and the sum drives \(h_3[n]\). With \(h_1[n]=\delta[n-1]\), \(h_2[n]=2\delta[n-2]\) and \(h_3[n]=\left(\tfrac13\right)^{n}u[n]\), find the overall impulse response and test it for stability.
- A system with \(h_1(t)=e^{-t}u(t)\) is cascaded with a pure delay of 2 seconds. Write the overall impulse response, and state whether the result depends on which block comes first.
- Show that convolving any signal with \(\delta(t-t_1)*\delta(t-t_2)\) delays it by \(t_1+t_2\), and use this to explain why delays in a cascade simply add.
- Classify each of the following as causal or non-causal, and stable or unstable: \(h_1(t)=e^{-3t}u(t-1)\); \(h_2(t)=e^{3t}u(-t)\); \(h_3[n]=\left(\tfrac54\right)^{n}u[-n]\); \(h_4[n]=\delta[n+4]\).
- For what real values of \(a\) is the system with \(h[n]=a^{n}u[n] + 3^{n}u[-n-1]\) BIBO stable? Treat the two terms separately, and be careful about which direction each one decays in.
- A system has \(h(t)=\dfrac{\sin t}{t}\). Show that \(h(t)\to 0\) as \(t\to\infty\) but that the system is not BIBO stable, and say what property of \(h\) is responsible.
- Prove the discrete-time BIBO condition \(\sum_k |h[k]| \lt \infty\) is necessary by constructing the bounded input \(x[n]=\operatorname{sgn}(h[-n])\) and evaluating \(y[0]\).
- The step response of a discrete-time system is \(s[n]=\big(1-\left(\tfrac12\right)^{n+1}\big)u[n]\). Find \(h[n]\), verify that \(\sum_n h[n]=s[\infty]\), and find the response to \(x[n]=u[n]-u[n-3]\).