- How to read the convolution sum — what the reversed index \(n-k\) is actually doing, and why reversal is forced on us rather than chosen.
- The four-step recipe — flip, shift, multiply, sum — and how to execute it graphically without losing track of which variable is which.
- Two shortcuts that make finite-length discrete convolution almost instant: the multiplication table and the length rule \(N+M-1\).
- Why the continuous-time integral is easy and its limits are hard, and the region-by-region method that gets them right every time.
- The two arithmetic checks — the area rule and the endpoint rule — that catch most convolution errors before they reach the answer.
- Convolution with impulses and steps: \(x*\delta(t-t_0) = x(t-t_0)\), and the step response as the running integral of \(h\).
- A short catalogue of standard convolutions — rectangles, exponentials, geometric sequences — worth committing to memory before an examination.
What Convolution Actually Asks
Chapter 6 ended with a formula and a promise. The formula was \(y[n] = \sum_k x[k]h[n-k]\); the promise was that it answers every question about an LTI system. What it did not do was tell you how to evaluate it, and the expression as written is easy to misread. This chapter is about execution.
Start by reading the sum out loud in the language of its derivation. The input \(x[k]\) is a list of impulse weights. Each weight launches a copy of the impulse response, delayed to start at \(k\). At the observation time \(n\), the copy launched at \(k\) has been running for \(n-k\) samples, so it currently contributes \(x[k]\,h[n-k]\). The output at time \(n\) is the total of every copy still contributing.
So the reversal is not a convention someone imposed. It arises because two clocks are running at once: \(k\) marks when a contribution was launched, \(n\) marks when we look, and what matters to \(h\) is the difference. Whenever the two clocks appear together in a single expression, one of them must run backwards relative to the other.
Before computing anything, one property is worth having in hand. Substituting \(m = n-k\) in the sum, so that \(k = n-m\) and \(m\) runs over all integers as \(k\) does, turns \(\sum_k x[k]h[n-k]\) into \(\sum_m h[m]x[n-m]\). The two forms are equal.
A change of dummy variable, nothing more. Practically it means you may reverse and slide either signal — always choose the simpler one, which is usually the shorter, or the one without a discontinuity. Chapter 8 explores what this symmetry means about systems; here it is a labour-saving device.
Flip, Shift, Multiply, Sum
Every convolution, in either domain, is the same four operations. Fix an output index \(n\) and treat it as a constant while you work. All plotting is done against the dummy variable \(k\) — this is the single point at which most errors are made, because the natural instinct is to keep plotting against \(n\).
Step 2 deserves care. To get \(h[n-k]\) from \(h[-k]\) you shift by \(n\) to the right when \(n\) is positive, because \(h[n-k] = h[-(k-n)]\) — the reversed signal has been moved so that its reference point sits at \(k=n\). A useful anchor: the sample of \(h\) that was at \(k=0\) before flipping ends up at \(k=n\) after both operations. Locate that one sample and the whole reversed sequence is placed.
Step 4 produces one number. To get the whole output signal you repeat steps 2 to 4 for every \(n\), which is why the practical skill is not doing one sum but recognising the ranges of \(n\) over which the sum has the same form. For finite-length signals those ranges are easy: the sum is empty until the two sequences start to overlap and empty again once they have passed each other.
Graphical Convolution in Discrete Time
Take \(x[n] = \{1, 2, 3\}\) for \(n = 0,1,2\) and \(h[n] = \{3, 1, 2\}\) for \(n = 0,1,2\), both zero elsewhere, and carry out the recipe. The first two panels below are the signals plotted against \(k\); the third is the reversed and shifted impulse response for the particular choice \(n = 2\), drawn on top of \(x[k]\) so the overlapping products can be read off.
Check the third panel against the recipe. Flipping \(h[k] = \{3,1,2\}\) gives \(h[-k]\), which places 3 at \(k=0\), 1 at \(k=-1\) and 2 at \(k=-2\). Shifting right by \(n=2\) moves the value 3 to \(k=2\), the 1 to \(k=1\) and the 2 to \(k=0\). So the reversed-and-shifted sequence, read left to right from \(k=0\), is \(\{2, 1, 3\}\) — which is what the gold stems show. Multiplying against \(x[k] = \{1,2,3\}\) and adding gives \(1\cdot2 + 2\cdot1 + 3\cdot3 = 13\), so \(y[2] = 13\).
Now run through the other values of \(n\). For \(n \lt 0\) the reversed sequence lies entirely to the left of \(x\) and there is no overlap, so \(y[n] = 0\). At \(n=0\) only one pair overlaps, at \(n=1\) two pairs, at \(n=2\) all three, and thereafter the overlap shrinks again until at \(n=5\) it is gone. Writing out the four remaining sums:
The complete output is \(y[n] = \{3, 7, 13, 7, 6\}\) for \(n = 0\) to \(4\). Two things are worth noticing before moving on. First, the answer is five samples long although both inputs were three — convolution spreads a signal out, and Section 7-4 turns that observation into a rule. Second, the sum of the output samples is \(3+7+13+7+6 = 36\), which is exactly \((1+2+3)(3+1+2) = 6 \times 6\). That is not a coincidence either.
Total areas multiply. Interchanging the order of summation proves it in one line: \(\sum_n\sum_k x[k]h[n-k] = \sum_k x[k]\sum_n h[n-k]\), and the inner sum is the total area of \(h\) whatever \(k\) is. Use it on every problem — it costs seconds and catches most arithmetic slips.
Two Shortcuts for Finite Sequences
Graphical convolution builds understanding but is slow. For finite-length sequences two shortcuts make the calculation almost mechanical, and both are worth having ready under examination conditions.
The first is the multiplication table. Write \(x\) across the top and \(h\) down the side, fill each cell with the product of its row and column entries, then add along the anti-diagonals. Each anti-diagonal is one output sample, in order.
| × | \(x[0]=1\) | \(x[1]=2\) | \(x[2]=3\) |
|---|---|---|---|
| \(h[0]=3\) | 3 | 6 | 9 |
| \(h[1]=1\) | 1 | 2 | 3 |
| \(h[2]=2\) | 2 | 4 | 6 |
Read the anti-diagonals from the top-left corner outwards: \(3\); then \(6+1 = 7\); then \(9+2+2 = 13\); then \(3+4 = 7\); then \(6\). The result \(\{3,7,13,7,6\}\) agrees with the graphical calculation, obtained in a fraction of the time.
The reason this works is that convolution is polynomial multiplication. Associate with each sequence a polynomial whose coefficients are its samples — using \(z^{-1}\) for a delay, which is exactly the association Chapter 23 will formalise as the \(z\)-transform:
Collecting like powers in a polynomial product is precisely adding the anti-diagonals, and the power of \(z^{-1}\) is the sample index. This identity is the seed of one of the most useful facts in the whole subject: convolution in time becomes multiplication in a transform domain. Everything from Chapter 13 onwards is that sentence, generalised.
The second shortcut is the length rule, which tells you where the answer lives before you compute a single product.
If \(x\) occupies \(N\) samples and \(h\) occupies \(M\), the convolution occupies \(N+M-1\) samples, beginning where the two starting indices add. The end index likewise adds: \(n_{\text{end}}(y) = n_{\text{end}}(x) + n_{\text{end}}(h)\).
The rule follows straight from the overlap argument. The earliest \(n\) at which anything overlaps is the one where the first sample of the flipped \(h\) meets the first sample of \(x\), which happens at \(n = n_{\text{start}}(x) + n_{\text{start}}(h)\); the last is where the final samples meet. Counting inclusively between them gives \(N+M-1\), and the "\(-1\)" is simply because both ends are counted.
Together with the area rule this gives three independent checks on any finite convolution: the first sample must equal the product of the two first samples, the last must equal the product of the two last samples, and the total must equal the product of the two totals. In the worked case: \(3 = 1\times3\) ✓, \(6 = 3\times2\) ✓, \(36 = 6\times6\) ✓.
The Convolution Integral
Continuous time uses the same four steps with sums replaced by integrals and stems replaced by areas. The integrand is rarely the difficulty — it is usually a product of exponentials or a constant. The difficulty is the limits, because the range over which the two signals overlap changes as \(t\) changes, and the answer therefore has different formulas on different intervals of \(t\).
Everything is plotted against \(\tau\), and \(t\) is a constant during the integration — a slider you set before you begin. The signal \(h(t-\tau)\), read as a function of \(\tau\), is \(h\) reversed and then translated so that its argument-zero point sits at \(\tau = t\). If \(h\) is causal, so that \(h(\tau) = 0\) for \(\tau \lt 0\), then \(h(t-\tau)\) is non-zero only for \(\tau \lt t\): the reversed impulse response occupies everything to the left of the slider.
That single observation collapses the general integral for the most common case in engineering. If both signals are causal, the integrand vanishes unless \(0 \lt \tau \lt t\), and the infinite limits become finite ones.
The lower limit comes from \(x(\tau)=0\) for \(\tau \lt 0\); the upper limit from \(h(t-\tau)=0\) for \(\tau \gt t\). If \(t\) is negative the two constraints are contradictory, the interval is empty, and the output is zero — as it must be for a causal system driven by a signal that has not started.
With those limits fixed, exponential convolutions become routine. Take \(x(t) = e^{-a t}u(t)\) and \(h(t) = e^{-b t}u(t)\) with \(a \ne b\). For \(t \ge 0\),
The trick that made it work is pulling \(e^{-bt}\) out of the integral, which is legal because the integration variable is \(\tau\). What remains is an elementary exponential integral. Two sanity checks: at \(t=0\) the answer is zero, as it must be since the interval of integration has collapsed; and as \(t\) grows both terms decay, with the slower one — the smaller of \(a\) and \(b\) — eventually dominating. The result is a signal that rises from zero, peaks, and decays with the slower of the two time constants, which is the characteristic shape of two first-order stages in cascade.
The case \(a = b\) cannot be obtained by substitution because the denominator vanishes; it must be done separately. With \(b=a\) the integrand becomes \(e^{-a\tau}e^{-a(t-\tau)} = e^{-at}\), independent of \(\tau\), so the integral is just \(e^{-at}\) times the length of the interval:
Getting the Limits Right: Convolution by Regions
When the signals have finite duration, the overlap does not simply grow — it grows, then saturates, then shrinks. Each phase gives a different formula, and the whole art is in identifying the phases and their boundaries. The method is always the same: slide the reversed signal from \(-\infty\) to \(+\infty\) and note every value of \(t\) at which an edge of one signal crosses an edge of the other. Those crossings are the region boundaries.
Work it through on the standard case. Let \(x(t)\) be a unit-height pulse on \(0 \lt t \lt 2\) and \(h(t)\) a unit-height pulse on \(0 \lt t \lt 3\). Plotted against \(\tau\), the signal \(x(\tau)\) is fixed on \([0,2]\) and the reversed pulse \(h(t-\tau)\) occupies \([t-3,\;t]\), a window of width 3 whose right edge sits at the slider position \(t\).
Four boundaries, then, at \(t=0\), \(t=2\), \(t=3\) and \(t=5\), separating five regions — of which the outer two are trivially zero. Each boundary is an edge crossing: at \(t=0\) the right edge of the window reaches the left edge of \(x\); at \(t=2\) it reaches the right edge of \(x\); at \(t=3\) the window's left edge reaches the left edge of \(x\); at \(t=5\) it passes the right edge and overlap ends. Because both signals have unit height, the integral in each region is simply the length of the overlap.
Apply the checks. The result runs from \(t=0\) to \(t=5\), and \(0+0=0\), \(2+3=5\) — the continuous-time version of the length rule, where durations add. The area under the trapezoid is \(\tfrac12(2)(2) + (1)(2) + \tfrac12(2)(2) = 2+2+2 = 6\), and the areas of the two rectangles are 2 and 3, whose product is 6. Both checks pass.
Two special cases are worth recording. If the rectangles have equal width \(T\), the flat region has zero length and the trapezoid degenerates into a triangle of base \(2T\) and peak \(T\) at \(t=T\). And whichever rectangle is wider, the flat region has length equal to the difference of the widths — so the shape of the answer immediately tells you how unequal the inputs were.
Impulses, Steps and the Step Response
Two convolutions occur so often that computing them from scratch is a waste of effort. Both follow directly from Chapter 3's impulse properties.
The first is convolution with an impulse. Sifting gives it immediately: \(\int x(\tau)\delta(t-t_0-\tau)\,d\tau\) reads off \(x\) at the point where the impulse's argument vanishes, namely \(\tau = t-t_0\).
The impulse is the identity element for convolution. A system with \(h(t) = \delta(t)\) is a plain wire; one with \(h(t) = \delta(t-t_0)\) is a pure delay. This is why the impulse response of any system built from adders, gains and delays can be written down by inspection.
Because convolution is linear in each argument, this extends to sums of impulses. Convolving with \(\sum_k a_k\delta(t - t_k)\) produces \(\sum_k a_k x(t-t_k)\) — the signal replicated at every impulse location and scaled by the weights. The special case in which the impulses form a uniform train, \(\sum_k \delta(t-kT)\), replicates the signal periodically, and that observation is the whole mechanism behind sampling and the spectral repetition of Chapter 20.
The second standard convolution is with the unit step. Since \(u(t-\tau) = 1\) for \(\tau \lt t\) and 0 beyond, the step simply truncates the integral:
Applied to \(h\) itself, this defines the quantity a laboratory most often measures, because a step is far easier to generate than an impulse. The step response \(s(t)\) is the output when the input is \(u(t)\).
Each determines the other, so measuring either one characterises the system completely. The pair is the singularity-family relation of Chapter 3 seen through a system.
An example makes the relation concrete. For \(h(t) = e^{-2t}u(t)\), integrating from 0 to \(t\) gives \(s(t) = \tfrac12\big(1-e^{-2t}\big)u(t)\) — the familiar charging curve, rising from zero towards the final value \(\tfrac12\), which is the total area under \(h\). Differentiating \(s\) recovers \(e^{-2t}u(t)\), as it must, and the fact that \(s\) settles to a finite value rather than growing without bound is exactly the statement that \(h\) is absolutely integrable. Chapter 8 turns that remark into the stability criterion.
A Working Catalogue
A small number of convolutions recur constantly, in this book and in examinations. Deriving each one takes a few minutes; recognising it takes a few seconds. The results below are all obtainable by the methods of the preceding sections, and it is worth deriving each once and then memorising it.
| \(x\) | \(h\) | \(x*h\) |
|---|---|---|
| \(x(t)\) | \(\delta(t-t_0)\) | \(x(t-t_0)\) |
| \(x(t)\) | \(u(t)\) | \(\displaystyle\int_{-\infty}^{t}x(\tau)\,d\tau\) |
| \(u(t)\) | \(u(t)\) | \(t\,u(t) = r(t)\) |
| \(e^{-at}u(t)\) | \(u(t)\) | \(\dfrac{1-e^{-at}}{a}u(t)\) |
| \(e^{-at}u(t)\) | \(e^{-bt}u(t)\) | \(\dfrac{e^{-at}-e^{-bt}}{b-a}u(t)\), \(a \ne b\) |
| \(e^{-at}u(t)\) | \(e^{-at}u(t)\) | \(t\,e^{-at}u(t)\) |
| pulse, width \(T\) | pulse, width \(T\) | triangle, base \(2T\), peak \(T\) |
| pulse, width \(T_1\) | pulse, width \(T_2\) | trapezoid, base \(T_1+T_2\), flat top of length \(|T_1-T_2|\) |
| \(x[n]\) | \(\delta[n-m]\) | \(x[n-m]\) |
| \(u[n]\) | \(u[n]\) | \((n+1)u[n]\) |
| \(\alpha^n u[n]\) | \(u[n]\) | \(\dfrac{1-\alpha^{\,n+1}}{1-\alpha}u[n]\), \(\alpha \ne 1\) |
| \(\alpha^n u[n]\) | \(\alpha^n u[n]\) | \((n+1)\alpha^n u[n]\) |
| \(\alpha^n u[n]\) | \(\beta^n u[n]\) | \(\dfrac{\alpha^{\,n+1}-\beta^{\,n+1}}{\alpha-\beta}u[n]\), \(\alpha \ne \beta\) |
The parallels between the two halves of the table are not accidental. Each discrete-time entry is the continuous-time entry with the integral replaced by a geometric sum, and each degenerate case — \(a=b\), \(\alpha=\beta\), \(\alpha=1\) — produces the same characteristic extra factor of \(n\) or \(t\). That factor is the signature of a repeated root, and Chapter 9 will meet it again when solving differential equations whose characteristic polynomial has a double root. Two entirely different calculations produce the same \(t\,e^{-at}\) because they are, underneath, the same calculation.
Worked Examples
Problem. Convolve \(x[n] = \{1, -1, 2\}\) for \(n = -1, 0, 1\) with \(h[n] = \{2, 3\}\) for \(n = 0, 1\).
Solution. First locate the answer with the length rule. Lengths 3 and 2 give \(3+2-1 = 4\) samples; the start indices \(-1\) and \(0\) add to \(-1\); the end indices \(1\) and \(1\) add to \(2\). So the result occupies \(n = -1\) to \(n = 2\), four samples — consistent.
Now use the multiplication table, remembering that the table itself is blind to indexing — it produces the sample values in order, and the length rule tells us where to place them.
| × | 1 | −1 | 2 |
|---|---|---|---|
| 2 | 2 | −2 | 4 |
| 3 | 3 | −3 | 6 |
Anti-diagonals from the top-left: \(2\); \(-2+3 = 1\); \(4-3 = 1\); \(6\). Placing them from \(n=-1\),
Check by the area rule: \(\sum x = 1-1+2 = 2\), \(\sum h = 2+3 = 5\), and \(\sum y = 2+1+1+6 = 10 = 2 \times 5\) ✓. Check the endpoints: \(y[-1] = 1\times2 = 2\) ✓ and \(y[2] = 2\times3 = 6\) ✓.
Problem. Find \(y[n] = x[n]*h[n]\) for \(x[n] = u[n]\) and \(h[n] = (1/2)^n u[n]\), and state the final value.
Solution. Write the sum and use the two step functions to fix the limits. The term \(u[k]\) forces \(k \ge 0\) and \(h[n-k]\) forces \(n-k \ge 0\), that is \(k \le n\). For \(n \lt 0\) these are contradictory and \(y[n] = 0\); for \(n \ge 0\),
where the substitution \(m = n-k\) simply reverses the order of the terms. So \(y[n] = \big(2 - (1/2)^n\big)u[n]\), giving \(y[0] = 1\), \(y[1] = 1.5\), \(y[2] = 1.75\), and a limit of 2 as \(n\) grows.
The checks are instructive. Directly, \(y[0]\) must be \(x[0]h[0] = 1\) ✓ and \(y[1] = h[0]+h[1] = 1.5\) ✓. The final value 2 is the total area \(\sum_k h[k] = 1/(1-\tfrac12) = 2\), which is what convolving with a step must give: a running sum that eventually accumulates everything. This is the discrete-time step response of the system, and its convergence to a finite number is the stability condition of Chapter 8 in visible form.
Problem. An LTI system with \(h(t) = e^{-3t}u(t)\) is driven by \(x(t) = e^{-2t}u(t)\). Find \(y(t)\), its peak value and the time at which the peak occurs. Then state what changes if the driving exponential is also \(e^{-3t}\).
Solution. Both signals are causal, so the limits are 0 to \(t\), and the catalogue entry with \(a=2\), \(b=3\) applies:
For the peak, differentiate and set to zero: \(-2e^{-2t}+3e^{-3t} = 0\), so \(e^{t} = 3/2\) and \(t_{\text{peak}} = \ln 1.5 = 0.405\) s. Substituting back,
using \(e^{-t_{\text{peak}}} = 2/3\). The curve starts at zero, rises to \(4/27\) at \(t = 0.405\), and then decays; for large \(t\) the \(e^{-2t}\) term dominates, so the tail is governed by the slower of the two exponentials. Area check: \(\int y\,dt = \tfrac12 - \tfrac13 = \tfrac16\), and the areas of \(x\) and \(h\) are \(\tfrac12\) and \(\tfrac13\), whose product is \(\tfrac16\) ✓.
If instead \(x(t) = e^{-3t}u(t)\), the formula cannot be used — its denominator \(b-a\) vanishes. Returning to the integral, the integrand becomes \(e^{-3\tau}e^{-3(t-\tau)} = e^{-3t}\), a constant in \(\tau\), so \(y(t) = t\,e^{-3t}u(t)\). Matched rates give a linear-times-exponential response rather than a difference of two exponentials, and it is a different shape: still zero at the origin, but peaking at \(t = 1/3\) s, since \(\tfrac{d}{dt}\big(te^{-3t}\big) = e^{-3t}(1-3t)\) vanishes there.
Problem. A system has \(h(t) = e^{-t}u(t)\). Find its response to the pulse \(x(t) = u(t)-u(t-2)\), by regions.
Solution. Convolve with \(x\) flipped, which is easier here because \(x\) is the piecewise-constant one: \(y(t) = \int h(\tau)x(t-\tau)\,d\tau\). The reversed pulse \(x(t-\tau)\) equals 1 for \(t-2 \lt \tau \lt t\) and zero elsewhere, so the integrand is \(e^{-\tau}\) over the intersection of that window with \(\tau \gt 0\).
Region 1, \(t \lt 0\). The window lies entirely in \(\tau \lt 0\), where \(h\) vanishes. \(y(t) = 0\).
Region 2, \(0 \le t \le 2\). The window's left edge is still negative, so the intersection is \(0 \lt \tau \lt t\):
Region 3, \(t \gt 2\). Now the whole window sits in \(\tau \gt 0\), so the intersection is \(t-2 \lt \tau \lt t\):
The two expressions agree at \(t=2\): region 2 gives \(1-e^{-2} = 0.8647\), and region 3 gives \(e^{-2}(e^2-1) = 1-e^{-2} = 0.8647\) ✓. Continuity at every region boundary is a required check — a mismatch always means a limit was set wrongly. Physically this is a capacitor charging while the pulse is present and discharging once it is removed, and the peak occurs exactly at the falling edge.
Problem. For \(h(t) = e^{-2t}u(t)\), find (a) the step response \(s(t)\); (b) the response to \(x(t) = \delta(t-1) - 2\delta(t-3)\); (c) the response to the pulse \(u(t)-u(t-2)\), using \(s(t)\).
Solution (a). Integrate \(h\) from 0 to \(t\):
rising from 0 towards \(\tfrac12\), which is the total area under \(h\). Differentiating returns \(e^{-2t}u(t)\) ✓.
Solution (b). Convolution with an impulse is a shift, and convolution is linear, so no integration is needed at all:
a decaying exponential starting at \(t=1\), joined at \(t=3\) by a second, inverted, twice as large. At \(t=3\) the first has fallen to \(e^{-4} = 0.0183\) while the second starts at \(-2\), so the output jumps downwards to \(-1.982\) there.
Solution (c). Since the pulse is \(u(t)-u(t-2)\) and the system is LTI, the response is the same combination of step responses:
For \(0 \le t \le 2\) this is \(\tfrac12(1-e^{-2t})\), and for \(t \gt 2\) the two constants cancel, leaving \(\tfrac12\big(e^{-2(t-2)}-e^{-2t}\big) = \tfrac12 e^{-2t}(e^{4}-1)\). Charging then discharging, as in Example 4, and obtained here without a single integral — a reminder that the step response, once known, answers every piecewise-constant input.
Problem. Find \(y[n] = x[n]*h[n]\) for \(x[n] = (1/3)^n u[n]\) and \(h[n] = u[n]-u[n-4]\), the four-point rectangular window.
Solution. Flip the window, since it is the finite one: \(y[n] = \sum_k x[k]\,h[n-k]\), and \(h[n-k] = 1\) exactly when \(0 \le n-k \le 3\), that is \(n-3 \le k \le n\). Combined with \(x[k] = 0\) for \(k \lt 0\), the sum runs over \(\max(0, n-3) \le k \le n\). Three regions follow.
Region 1, \(n \lt 0\). Empty range, \(y[n] = 0\).
Region 2, \(0 \le n \le 3\). The window is still partly hanging off the left end of \(x\), so the sum starts at \(k=0\):
giving \(y[0]=1\), \(y[1]=4/3\), \(y[2]=13/9\) and \(y[3]=40/27\).
Region 3, \(n \ge 4\). The window now lies entirely inside \(x\), and always covers exactly four terms:
Check the join: region 2 at \(n=3\) gives \(\tfrac32\big[1-(1/3)^4\big] = \tfrac32\cdot\tfrac{80}{81} = \tfrac{40}{27}\), and region 3 at \(n=3\) gives \(\tfrac{40}{27}(1/3)^0 = \tfrac{40}{27}\) — the two formulas agree at the boundary, as they must.
The steady behaviour is worth rewriting. Since \((1/3)^{n-3} = 27\,(1/3)^{n}\), region 3 says simply \(y[n] = 40\,(1/3)^{n}\) for \(n \ge 3\): after the window has fully entered, the output is the input multiplied by a constant gain of 40. That is the eigenfunction property of Chapter 6 in action, and the gain is the eigenvalue. Evaluating \(H(z) = \sum_k h[k]z^{-k}\) for this four-point window at \(z = 1/3\) gives \(1 + 3 + 9 + 27 = 40\) — exactly the observed gain. The three-sample transient at the start is the price of the input having been switched on at \(n=0\) rather than having run forever, which is what the eigenfunction argument assumes.
Chapter Summary
Flip, shift, multiply, sum — all plotted against the dummy variable \(k\) or \(\tau\), with \(n\) or \(t\) held fixed as a slider.
Either signal may be reversed. Always flip the shorter or simpler one; it is a change of dummy variable, not a new result.
The multiplication table gives finite convolutions in seconds; the length rule \(N+M-1\) says where the answer lives before you start.
Continuous-time difficulty is entirely in the limits. Boundaries occur where edges cross; formulas must agree at each boundary.
Areas multiply, durations add, endpoints are products of endpoints, and adjacent regions must join continuously.
\(x*\delta(t-t_0) = x(t-t_0)\); \(x*u\) is the running integral; \(s(t)=\int_{-\infty}^{t}h\), and \(h = ds/dt\).
Practice Problems
Problems 1 to 3 are discrete-time drill; 4 to 6 are continuous-time convolutions by regions; 7 and 8 are about the structure of the operation rather than its arithmetic. Apply the area and endpoint checks to every numerical answer before moving on.
- Convolve \(x[n] = \{2, 1, -1, 3\}\) at \(n = 0\ldots3\) with \(h[n] = \{1, 2\}\) at \(n = 0,1\), using the multiplication table. Verify the length, the two endpoints and the total area.
- Convolve \(x[n] = \{1, 0, 2\}\) at \(n = -2,-1,0\) with \(h[n] = \{3, -1, 1\}\) at \(n = 1,2,3\). State the index range of the result before computing it.
- Find \(y[n] = (0.8)^n u[n] * u[n]\) in closed form, and evaluate its limit as \(n \to \infty\). Explain that limit in terms of the total area of \((0.8)^n u[n]\).
- Two pulses have unit height: \(x(t)\) on \(0 \lt t \lt 1\) and \(h(t)\) on \(0 \lt t \lt 4\). Find and sketch \(x*h\) by regions, marking every boundary and checking the total area.
- A system has \(h(t) = e^{-5t}u(t)\). Find its response to \(x(t) = 3\big[u(t-1)-u(t-4)\big]\), giving a separate expression for each region and verifying continuity at both boundaries.
- Evaluate \(x(t)*h(t)\) for \(x(t) = e^{2t}u(-t)\) and \(h(t) = e^{-t}u(t)\). Note carefully that \(x\) is anti-causal, so the limits of Section 7-5 do not apply; work out the overlap directly and expect two regions.
- A system's step response is measured to be \(s(t) = \big(1-e^{-4t}\big)u(t)\). Find \(h(t)\), then find the response to \(x(t) = u(t) - 2u(t-1) + u(t-2)\) without evaluating any integral.
- Prove the area rule \(\int y \,dt = \big(\int x\,dt\big)\big(\int h\,dt\big)\) by interchanging the order of integration, and state the condition on \(x\) and \(h\) under which the interchange is justified. Then use it to show that a system whose \(h\) has zero total area must produce zero total area at its output for every input.