- Why linearity and time invariance together are worth a whole part of this book, when neither alone gives much leverage.
- Superposition in its working form — extended from two terms to an infinite sum — and the quiet assumption that extension involves.
- The impulse response \(h[n]\) and \(h(t)\): a complete description of an LTI system written as a single signal.
- A full derivation of the convolution sum \(y[n]=\sum_k x[k]h[n-k]\) from the impulse decomposition of Chapter 3, in four lines.
- The staircase argument that turns that sum into the convolution integral \(y(t)=\int x(\tau)h(t-\tau)\,d\tau\).
- Why \(e^{st}\) and \(z^n\) are eigenfunctions of every LTI system, and how that single fact creates the Laplace, Fourier and \(z\)-transforms of Parts 3 to 7.
- What the LTI assumption excludes — modulators, samplers, squarers, biased amplifiers — and how engineers keep using it anyway.
Why This One Class Deserves a Whole Part
Chapter 5 laid out six properties a system might have and tested a gallery of examples against each. That was taxonomy. Now comes the reason for the taxonomy: two of those six properties, and only those two, are strong enough to make a general theory possible.
Consider what "solving" a system means in the general case. A system is a rule that maps an entire input signal to an entire output signal. To know a general system completely you would have to be told its output for every conceivable input — an uncountable catalogue, and no more useful than the system itself. There is no shortcut, no formula, no compact description. This is the ordinary situation, and it is hopeless.
Now impose linearity. Suddenly the catalogue can shrink, because knowing the outputs for a set of basis signals tells you the outputs for every combination of them. Impose time invariance as well, and the catalogue collapses further still: you no longer need one entry per basis signal, because the basis signals are all time shifts of one another and the system treats shifts as shifts. Between them, the two properties reduce an infinite catalogue to one entry.
That one entry is the impulse response. This chapter earns it. Chapter 3 supplied the necessary ingredient — the fact that any signal is a sum of scaled, shifted impulses — and left the promise that a system's answer to one impulse would determine its answer to everything. Here that promise is kept, and the result is the operation the rest of Part 2 revolves around.
Linearity, Put to Work
Write \(T\{\cdot\}\) for the system, so that \(y = T\{x\}\) means "\(y\) is what the system produces when \(x\) goes in". Chapter 5 defined the system to be linear when scaling and addition pass straight through it.
The combined form is the one to remember, and it extends immediately. Apply it to three terms by grouping two of them, then four, then any finite number, and induction gives superposition in the shape we actually want to use it.
Decompose the input into pieces you can handle, find the response to each piece separately, then add the responses with the same coefficients. This is the "decompose, respond, superpose" strategy of Chapter 1, and linearity is precisely the licence to use it.
Two consequences follow at once and both are used as quick tests. First, setting \(a = 0\) gives \(T\{0\} = 0\): a linear system at rest with no input must produce no output. A system whose output contains a constant term regardless of input — an amplifier with a DC offset, say — cannot be linear. Second, the response to \(-x\) is \(-y\), so a linear system cannot rectify, clip or saturate, since all three treat positive and negative inputs differently.
The first test disposes of a common family of systems in one line, and it is worth naming the family. A system such as \(y[n] = 2x[n] + 5\) obeys neither additivity nor homogeneity, yet it is clearly not "nonlinear" in any troubling sense: it is a linear system with a constant added. Such systems are called incrementally linear — the response to a change in the input is linear even though the response to the input is not. The standard repair is to work with deviations from an operating point, which is how a transistor amplifier, thoroughly nonlinear as a device, is analysed with linear tools.
Time Invariance, Put to Work
The second property says the system does not care what time it is. Run an experiment today and repeat it tomorrow with the same input, and you get the same output, merely later. Formally, delaying the input delays the output by the same amount and changes nothing else.
The clean way to picture this is as a pair of paths that must agree. Shifting and then processing must give the same signal as processing and then shifting — the delay block and the system block commute. Every practical test for time invariance is this diagram written out algebraically: compute the top path, compute the bottom path, compare.
Chapter 5 showed the test in action, and the informative failures are worth recalling because they recur throughout this book. The modulator \(y(t) = x(t)\cos(\omega_c t)\) is perfectly linear yet fails, because the carrier is nailed to the time origin: shifting the input does not shift the carrier with it. The compressor \(y[n] = x[2n]\) is linear and fails as well, because a shift of one sample at the input emerges as a shift of half a sample at the output, which is not a shift at all. Both of these are useful, everyday systems — they are simply outside the theory this chapter builds, and Parts 6 and 7 treat them with tools of their own.
The practical signature of time invariance is that the coefficients describing the system do not depend on \(t\) or \(n\). A resistor whose value is fixed gives a time-invariant circuit; a resistor being turned by hand does not. That is exactly why Chapter 9 restricts itself to differential and difference equations with constant coefficients.
The Two Together
A system possessing both properties is called linear time-invariant, universally abbreviated LTI. Neither property alone does much: linearity without time invariance still leaves a different response for every instant, and time invariance without linearity forbids decomposition entirely. Together they produce a statement that is small to write and enormous in consequence.
Suppose you know the system's response to one particular signal \(p(t)\), and call it \(q(t)\). Time invariance tells you the response to \(p(t-t_0)\) for every \(t_0\) — it is \(q(t-t_0)\), free of charge. Linearity then tells you the response to any weighted combination of those shifts. So from one input-output pair you obtain the response to the entire family of scaled, shifted copies of \(p\).
Scaling and shifting the input scales and shifts the output identically, and the results add. Everything else in Part 2 is this statement applied to a well-chosen \(p\).
The question is which \(p\) to choose, and Chapter 3 answered it before the question was asked. If the entire signal set can be written as scaled, shifted copies of one signal, that signal is the right choice. Exactly one candidate has that property: the unit impulse, because \(x[n] = \sum_k x[k]\delta[n-k]\) says that every signal is already a weighted sum of shifted impulses, with the weights sitting there in plain sight as the signal's own samples.
The Impulse Response and the Convolution Sum
Define the impulse response to be what the system does to a single unit impulse applied at the origin:
It is one signal, obtained from one experiment. The claim of this section is that for an LTI system it is not merely a description but the description: the system contains no information that \(h\) does not already carry. Discrete time proves it in four lines, each line using exactly one property.
Read the ladder from top to bottom. The first rung is the definition of \(h\). The second uses time invariance: an impulse arriving \(k\) samples late produces the same response, \(k\) samples late. The third uses homogeneity, and it is legitimate because \(x[k]\) is a fixed number — the sample value at index \(k\) — not a function of the running variable \(n\). The fourth uses additivity across all \(k\) at once. The left-hand column of the last rung is the impulse decomposition of Chapter 3, which is to say it is simply \(x[n]\); the right-hand column is therefore \(y[n]\), and we have it.
The output of any discrete-time LTI system is the convolution of the input with the impulse response. Give me \(h[n]\) and I can compute the response to every input there is, without knowing anything else about the system.
It is worth saying plainly what has been achieved, because the algebra is short enough to hide it. A system is an object of infinite complexity in general — a map between function spaces. For the LTI class that object has been replaced by a single ordinary signal and a single arithmetic operation. Two systems with the same impulse response are the same system. A system is designed by choosing \(h\); it is identified in the laboratory by measuring \(h\); it is simulated by convolving with \(h\). Chapter 7 is devoted to computing that convolution, and Chapter 8 to reading the system's properties — memory, causality, stability, invertibility — directly off \(h\).
From Sum to Integral
Continuous time needs the same argument with one extra step, because there is no such thing as "the sample at \(t = \tau\)" to serve as a coefficient. The repair is to approximate first and take a limit at the end.
Define a narrow pulse \(\delta_\Delta(t)\) of width \(\Delta\) and height \(1/\Delta\) — the same pulse Chapter 3 used to build the impulse. Chop the time axis into intervals of width \(\Delta\) and approximate \(x(t)\) by a staircase that holds the value \(x(k\Delta)\) across the \(k\)-th interval. Each stair is a scaled, shifted copy of that pulse, so the staircase is a sum of exactly the kind the master property handles.
The factor \(\Delta\) is there because \(\delta_\Delta\) has height \(1/\Delta\), so multiplying by \(\Delta\) restores unit height and the stair really does sit at \(x(k\Delta)\). Now let \(\hat{h}_\Delta(t)\) be the system's response to the single pulse \(\delta_\Delta(t)\). The master property gives the response to the whole staircase immediately:
Now shrink \(\Delta\). Three things happen at once, and each is the limit of an ordinary object. The staircase \(\hat{x}\) tends to \(x\); the narrow pulse \(\delta_\Delta\) tends to the impulse, so its response \(\hat{h}_\Delta\) tends to the impulse response \(h\); and the sum over \(k\), with \(k\Delta \to \tau\) and \(\Delta \to d\tau\), tends to a Riemann integral. What is left is the continuous-time counterpart of the convolution sum.
Identical in structure to the sum: weight each shifted impulse response by the input's value at that instant, and accumulate. The integral is over \(\tau\); \(t\) is a parameter held fixed while the integral is done, and the answer is a function of it.
The same result can be reached in one line by starting from the sifting form of the impulse decomposition, \(x(t) = \int x(\tau)\delta(t-\tau)\,d\tau\), and applying the master property directly to the integral. That derivation is shorter but conceals the limit: it treats \(\delta\) as if it were an ordinary signal that could be fed into a system, which Chapter 3 warned is never quite true. The staircase argument is longer precisely because it is honest — every object in it is a genuine finite-height pulse until the very last step.
Exponentials as Eigenfunctions
Impulses are not the only useful decomposition, and the second one is arguably more powerful. Chapter 1 asserted that complex exponentials pass through LTI systems unchanged in shape. With the convolution integral in hand that assertion becomes a two-line proof.
Put \(x(t) = e^{st}\) into the convolution integral, with \(s = \sigma + j\omega\) any complex number. It is convenient to use the equivalent form of the integral in which the shift sits on \(h\) rather than on the input; the substitution \(\lambda = t-\tau\) turns one into the other, as Chapter 7 sets out in full:
The step that matters is pulling \(e^{st}\) out of the integral, and it is allowed because the integration variable is \(\tau\), not \(t\). What remains is an integral with no \(t\) in it at all: a plain complex number, depending only on \(s\) and on the system. Call it \(H(s)\).
The exponential comes out as the same exponential, multiplied by a complex constant. The system cannot change its shape, its rate of growth or its frequency — only its amplitude and phase.
The word eigenfunction is borrowed from linear algebra deliberately. An eigenvector of a matrix is a vector the matrix cannot rotate, only stretch; \(e^{st}\) is a signal an LTI system cannot reshape, only scale. And \(H(s)\), the stretch factor, is the eigenvalue. The parallel is exact, because a linear system is a linear operator on a space of signals and convolution is its matrix.
This is where the rest of the book comes from. If a signal can be written as a combination of exponentials, then by superposition the output is that same combination with each term multiplied by its own \(H\) — and convolution, an integral, has been replaced by multiplication, which is arithmetic. Fourier series (Chapter 10) writes periodic signals as sums of \(e^{j\omega_0 kt}\); the Fourier transform (Chapter 13) writes aperiodic signals as integrals of \(e^{j\omega t}\); the Laplace transform (Chapter 17) uses general \(e^{st}\) and the \(z\)-transform (Chapter 23) uses \(z^n\). Four transforms, one motivation, and it is the box above.
One special case deserves separate attention because it is what a laboratory actually measures. Set \(s = j\omega\), so the input is the pure sinusoid-carrying exponential \(e^{j\omega t}\), and write \(H(j\omega) = |H(j\omega)|e^{j\angle H(j\omega)}\). For a system with real-valued \(h(t)\), feeding in \(\cos\omega t = \tfrac12(e^{j\omega t} + e^{-j\omega t})\) and superposing the two conjugate responses gives a real sinusoid out.
The frequency survives untouched; only amplitude and phase change, and both are read off the single complex number \(H(j\omega)\). This is the entire content of the phrase "frequency response", and it is why a filter can be specified by a gain-versus-frequency curve. Chapter 15 develops it properly; here it is a corollary of two properties and one integral.
Scope and Limits of the LTI Assumption
Everything above rests on both properties holding exactly. It is worth being clear about what is lost when either fails, and about how large the LTI class really is.
Drop time invariance but keep linearity, and the impulse response does not vanish — it acquires a second argument. The response to an impulse applied at time \(\tau\) is now some \(h(t,\tau)\) that depends separately on when the impulse arrived and when you look, rather than only on the elapsed time \(t-\tau\). Superposition still applies, so a valid formula survives.
The formula is correct but far weaker: \(h(t,\tau)\) is a function of two variables, exponentials are no longer eigenfunctions, and no transform turns the integral into a product. Drop linearity instead and even this much is gone — there is no decomposition, no superposition, and each input must be handled on its own terms.
The table below sorts the systems this book keeps returning to. Reading it, notice how many familiar and useful blocks fall outside the LTI class.
| System | Linear | Time-invariant | Comment |
|---|---|---|---|
| \(y(t) = 3x(t)\) | Yes | Yes | LTI; \(h(t) = 3\delta(t)\) |
| \(y(t) = x(t-3)\) | Yes | Yes | LTI; \(h(t) = \delta(t-3)\) |
| \(y(t) = dx/dt\) | Yes | Yes | LTI; \(h(t) = \delta'(t)\), the doublet |
| \(y(t) = t\,x(t)\) | Yes | No | Gain depends on the clock |
| \(y(t) = x(t)\cos\omega_c t\) | Yes | No | Amplitude modulator — carrier is fixed to \(t=0\) |
| \(y[n] = x[2n]\) | Yes | No | Downsampler; a shift in becomes half a shift out |
| \(y[n] = x^2[n]\) | No | Yes | Squarer; generates new frequencies |
| \(y[n] = 2x[n] + 5\) | No | Yes | Incrementally linear; \(T\{0\} \ne 0\) |
| \(y[n] = n\,x[n]\) | Yes | No | Growing gain; unstable and time-varying |
Half the table fails. Yet the theory of this part is applied constantly to real systems that are, strictly speaking, in the failing half — and the reason is that the failures are usually gentle. A transistor amplifier is nonlinear, but around a fixed bias point the deviations obey a linear law to excellent accuracy, which is the small-signal model. A satellite is time-varying as it burns fuel, but its mass changes over hours while its attitude dynamics settle in seconds, so it is time-invariant over any interval that matters. Even a modulator becomes LTI if you agree to analyse it in a frame rotating with the carrier.
The engineering skill is knowing when the approximation is safe and when it is not. Drive an amplifier hard enough and it clips: the small-signal model gives no warning, because clipping is exactly the behaviour linearity forbids. That failure mode — a model that is excellent until it is catastrophically wrong — is worth remembering every time an LTI analysis is used on a physical device.
Worked Examples
Problem. Determine whether each system is linear, time-invariant, or both: (a) \(y(t) = t\,x(t)\); (b) \(y[n] = x[n] + 3\); (c) \(y[n] = x[n] - x[n-1]\).
Solution (a). For linearity, feed in \(a x_1 + b x_2\): the output is \(t\big(ax_1(t)+bx_2(t)\big) = a\,t x_1(t) + b\,t x_2(t)\), which is \(a y_1 + b y_2\). Linear. For time invariance, compare the two paths of the commuting diagram. Shift-then-operate gives \(t\,x(t-t_0)\). Operate-then-shift means replacing \(t\) by \(t-t_0\) everywhere in the output expression, giving \((t-t_0)\,x(t-t_0)\). These differ, so the system is time-varying. Linear, not LTI.
Solution (b). Zero input gives \(y[n] = 3 \ne 0\), so linearity fails at once. Time invariance holds: both paths give \(x[n-n_0]+3\). It is incrementally linear — the difference between two outputs is \(x_1[n]-x_2[n]\), which is linear in the difference of the inputs.
Solution (c). Linearity is immediate since the expression is a weighted sum of input samples. For time invariance, shift-then-operate gives \(x[n-n_0] - x[n-n_0-1]\); operate-then-shift replaces \(n\) by \(n-n_0\) throughout, giving the same thing. LTI — and this is the first difference of Chapter 3, whose impulse response is \(h[n] = \delta[n]-\delta[n-1]\).
Problem. Find the impulse response of (a) \(y[n] = x[n] - 2x[n-1] + x[n-3]\); (b) the three-point moving average \(y[n] = \tfrac13\big(x[n]+x[n-1]+x[n-2]\big)\); (c) the continuous-time sliding integrator \(y(t) = \int_{t-T}^{t} x(\tau)\,d\tau\).
Solution (a) and (b). Both are already written as weighted sums of shifted inputs, so substituting \(x[n] = \delta[n]\) and using \(\delta[n-k]\) for each shifted term reads the answer straight off:
Equivalently \(h_a\) is the sequence \(\{1,-2,0,1\}\) starting at \(n=0\), and \(h_b\) is \(\{\tfrac13,\tfrac13,\tfrac13\}\). Whenever a system is described as a fixed linear combination of shifted inputs, its impulse response is nothing more than the list of coefficients.
Solution (c). Substitute \(x(\tau) = \delta(\tau)\). The integral \(\int_{t-T}^{t}\delta(\tau)\,d\tau\) equals 1 when the origin lies inside the interval of integration and 0 otherwise. The origin satisfies \(t-T \lt 0 \lt t\) exactly when \(0 \lt t \lt T\), so
a rectangular pulse of unit height and width \(T\). The system "remembers" exactly the last \(T\) seconds and weights them all equally, which is precisely what the shape of \(h_c\) says.
Problem. An LTI system has \(h[n] = \{1, 2, -1\}\) for \(n = 0,1,2\) and zero elsewhere. Find its response to \(x[n] = 2\delta[n] - \delta[n-1] + 3\delta[n-3]\).
Solution. The input is already a sum of scaled shifted impulses, so the master property applies directly — no convolution machinery is needed. Each impulse contributes a scaled, shifted copy of \(h\):
Now tabulate the three contributions. The term \(2h[n]\) puts \(\{2,4,-2\}\) at \(n=0,1,2\); the term \(-h[n-1]\) puts \(\{-1,-2,1\}\) at \(n=1,2,3\); the term \(3h[n-3]\) puts \(\{3,6,-3\}\) at \(n=3,4,5\). Adding column by column:
| \(n\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| \(2h[n]\) | 2 | 4 | −2 | 0 | 0 | 0 |
| \(-h[n-1]\) | 0 | −1 | −2 | 1 | 0 | 0 |
| \(3h[n-3]\) | 0 | 0 | 0 | 3 | 6 | −3 |
| \(y[n]\) | 2 | 3 | −4 | 4 | 6 | −3 |
So \(y[n] = \{2, 3, -4, 4, 6, -3\}\) for \(n = 0\) to \(5\). This table is convolution — Chapter 7 will formalise exactly this layout as the standard hand method.
Problem. The downsampler is \(y[n] = x[2n]\). Compute its impulse response, then compute its response to \(x[n] = \delta[n-1]\). Show that the convolution sum gives the wrong answer, and explain why.
Solution. Setting \(x[n] = \delta[n]\) gives \(h[n] = \delta[2n]\). Since \(2n = 0\) only when \(n=0\), and \(2n\) is never equal to any other integer for which \(\delta\) is non-zero, \(h[n] = \delta[n]\). By that measurement alone the system looks like a plain wire.
Now apply \(x[n] = \delta[n-1]\) directly. The output is \(y[n] = x[2n] = \delta[2n-1]\), and \(2n-1\) is odd for every integer \(n\), so it is never zero: \(y[n] = 0\) for all \(n\). The system swallows the input entirely.
What would convolution have predicted? With \(h[n] = \delta[n]\), the sum \(\sum_k \delta[k-1]\,h[n-k] = h[n-1] = \delta[n-1]\) — the input passed through unchanged. That is not the output. The prediction fails because the second rung of the ladder in Section 6-5 was time invariance, and this system does not have it: knowing the response to an impulse at \(n=0\) tells you nothing whatsoever about the response to an impulse at \(n=1\). The impulse response is a complete description only for LTI systems, and this example is the reason that qualifier is never dropped.
Problem. A causal LTI system has \(h(t) = e^{-2t}u(t)\). Find its response to \(x(t) = e^{j3t}\), and then to \(x(t) = \cos 3t\).
Solution. Because \(e^{st}\) is an eigenfunction, no convolution is required — only the eigenvalue. Evaluate \(H(s)\) at \(s = j3\):
The lower limit is 0 rather than \(-\infty\) because \(h(\tau)\) vanishes for \(\tau \lt 0\), and the upper limit converges provided \(\operatorname{Re}\{s\} \gt -2\), which \(s = j3\) satisfies. Hence
whose magnitude is \(1/\sqrt{13} = 0.2774\) and whose angle is \(-\arctan(3/2) = -56.31^\circ\). So the response to \(e^{j3t}\) is \(0.2774\,e^{j(3t - 56.31^\circ)}\).
For the real cosine, use the frequency-response corollary: the output is a cosine at the same frequency, scaled by \(|H(j3)|\) and delayed in phase by \(\angle H(j3)\),
Compare the effort: a convolution integral has been replaced by one substitution into a formula and a complex division. That saving, repeated across every frequency at once, is what Parts 3 to 5 are for.
Problem. An LTI system is known only through one experiment: when the input is \(x_1(t) = u(t)-u(t-1)\), the output \(y_1(t)\) is a triangle rising from 0 at \(t=0\) to 1 at \(t=1\) and falling back to 0 at \(t=2\). (a) Without finding \(h(t)\), determine the response to \(x_2(t) = u(t)-u(t-1) + 2\big[u(t-2)-u(t-3)\big]\). (b) Now find \(h(t)\) itself from the same single experiment.
Solution (a). Observe that \(x_2\) is built entirely from \(x_1\): the second bracket is \(2x_1(t-2)\), so \(x_2(t) = x_1(t) + 2x_1(t-2)\). The master property answers immediately, with no knowledge of \(h\) at all:
that is, the same triangle occupying \(0 \le t \le 2\), plus a triangle of height 2 occupying \(2 \le t \le 4\). They meet at \(t=2\), where both are zero, so the total is a triangle of height 1 followed by a triangle of height 2.
Solution (b). Use the same decomposition in the other direction. Since \(x_1(t) = u(t) - u(t-1)\), linearity and time invariance give \(y_1(t) = s(t) - s(t-1)\), where \(s\) is the system's step response. Differentiating, and using \(h = ds/dt\) — which holds because differentiation is itself LTI, so it commutes with the system —
The triangle has slope \(+1\) on \((0,1)\) and \(-1\) on \((1,2)\), so \(dy_1/dt = u(t) - 2u(t-1) + u(t-2)\). Matching term by term against \(h(t)-h(t-1)\) gives \(h(t) = u(t) - u(t-1)\): the system is the one-second sliding integrator of Example 2(c) with \(T=1\). The check is direct — a rectangle convolved with itself is a triangle, which Chapter 7 computes explicitly.
Chapter Summary
Linearity plus time invariance: scaled, shifted inputs give scaled, shifted outputs that add. Every result here follows from it.
\(h[n] = T\{\delta[n]\}\), \(h(t) = T\{\delta(t)\}\). For an LTI system it is a complete description — nothing else is needed.
\(y[n] = \sum_k x[k]h[n-k]\), derived in four rungs: definition, time invariance, scaling, additivity.
\(y(t) = \int x(\tau)h(t-\tau)d\tau\), obtained by shrinking the staircase approximation of the input to zero width.
\(e^{st} \to H(s)e^{st}\) and \(z^n \to H(z)z^n\). Convolution becomes multiplication, and every transform in this book follows.
Linear but time-varying systems keep a superposition integral with kernel \(h(t,\tau)\); nonlinear systems keep nothing.
Practice Problems
Problems 1 to 3 are property tests; 4 to 6 exercise the impulse response and the master property; 7 and 8 ask you to reproduce the derivations themselves, which is the surest way to own them.
- Test each system for linearity and time invariance, showing both paths of the commuting diagram where relevant: (a) \(y(t) = x(t/2)\); (b) \(y[n] = \sum_{k=-\infty}^{n} x[k]\); (c) \(y(t) = x(t)\,x(t-1)\); (d) \(y[n] = \cos\big(x[n]\big)\).
- Show that \(y[n] = 4x[n] - 7\) is not linear, then show that the system relating \(x_1[n]-x_2[n]\) to \(y_1[n]-y_2[n]\) is LTI, and give its impulse response.
- A system is described by \(y(t) = x(t)\cos(2\pi t)\). Verify that it is linear, then find a specific input and a specific shift for which the two paths of the commuting diagram disagree, and state the two outputs explicitly.
- Find the impulse response of (a) \(y[n] = x[n+1] - 2x[n] + x[n-1]\); (b) \(y(t) = \tfrac{1}{2}\big[x(t) + x(t-4)\big]\); (c) \(y(t) = \int_{-\infty}^{t} e^{-(t-\tau)}x(\tau)\,d\tau\).
- An LTI system has \(h[n] = \{3, -1, 2\}\) at \(n = 0,1,2\). Using the master property and a table like the one in Worked Example 3, find the response to \(x[n] = \delta[n] + 2\delta[n-2] - \delta[n-4]\).
- A continuous-time LTI system has \(h(t) = e^{-4t}u(t)\). Find \(H(j\omega)\), then compute the steady-state response to \(x(t) = 5\cos(3t) + 2\cos(8t)\), giving both amplitudes and both phase shifts in degrees. Comment on what the system does to the two components.
- Reproduce the derivation of the convolution sum from the impulse decomposition, stating at each of the four steps exactly which property is being used. Then explain where the derivation breaks down for the system \(y[n] = x[n]x[n-1]\), and where it breaks down for \(y[n] = x[2n]\).
- Starting from the staircase approximation, derive the convolution integral, and explain why the factor \(\Delta\) must appear in the expression for \(\hat{x}(t)\). Then show that letting \(x(t) = \delta(t)\) in the finished integral returns \(y(t) = h(t)\), as it must.