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Signals and Systems · 144 h · Topic 21 of 34

Convolution

Flip, shift, multiply and measure the area. With those four steps, and knowing the impulse response, you can compute the output of an LTI system for any input.

Convolution Impulse response Cascade Duration Discrete convolution

01Why the integral looks like this

Convolution is not a definition that fell from the sky: it comes from applying linearity and invariance to the decomposition of a signal into impulses. The reasoning was already sketched when discussing LTI systems and it is worth having in full:

\[ x(t) = \int x(\tau)\,\delta(t-\tau)\,d\tau \] \[ \delta(t-\tau) \;\xrightarrow{\ \text{invariance}\ }\; h(t-\tau) \] \[ \xrightarrow{\ \text{linearity}\ }\; y(t) = \int x(\tau)\,h(t-\tau)\,d\tau \]

Each instant of the input triggers a copy of \( h \), scaled by the value the input had at that instant and shifted to that point. The output is the superposition of all those copies. That is how the integral is read:

\[ y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t - \tau)\,d\tau \qquad\qquad y[n] = \sum_{k=-\infty}^{\infty} x[k]\,h[n-k] \]

The \( h(t-\tau) \) contains the two transformations already studied: the minus sign on \( \tau \) is a reflection and the \( t \) is a shift. That is why the graphical procedure is always the same: reflect, shift, multiply and measure the area.

02Reflect, shift, multiply, measure

For each value of \( t \), the integral is computed in four steps. The lab shows all four at once: above, the fixed signal and the reflected one sliding along, with the product shaded; below, the area of that shaded region drawn point by point, which is the output.

Lab · convolution, step by step

Above, \( x(\tau) \) in blue and \( h(t-\tau) \) in orange, reflected and shifted to the instant you choose. The green area is the product of the two. Below, that area as a function of \( t \): the output.

03Properties, and what they mean in a circuit

PropertyExpressionWhat it says about the circuit
Commutative\( x * h = h * x \) It makes no difference which one is reflected
Associative\( (x * h_1) * h_2 = x * (h_1 * h_2) \) Two stages in cascade are equivalent to one with \( h_1 * h_2 \)
Distributive\( x * (h_1 + h_2) = x*h_1 + x*h_2 \) Two branches in parallel add their responses
Identity element\( x * \delta = x \) An ideal wire has \( h = \delta \)
Shifting\( x * \delta(t-t_0) = x(t-t_0) \) A delay line
Duration\( T_y = T_x + T_h \) The output lasts longer than the input: the system “broadens”
Area\( A_y = A_x \cdot A_h \) The DC gain multiplies the area

The duration row explains something you can see on any oscilloscope: a narrow pulse that goes through a slow circuit comes out wider than it went in. And the area row is the time-domain version of the DC gain, because \( A_h = \int h(t)dt = H(0) \).

The cascade, and why Fourier is convenient

The associative property says that two filters in cascade are equivalent to a single one whose impulse response is \( h_1 * h_2 \). Computing that convolution by hand is tedious. What comes later in the subject is the shortcut: in the frequency domain convolution becomes a product, \( H(j\omega) = H_1(j\omega)H_2(j\omega) \), and putting stages in cascade becomes multiplying numbers. All the convenience of Fourier and Laplace is born here.

04The discrete version, which is just arithmetic

In discrete time there is no integral but a sum, and that makes it directly programmable:

\[ y[n] = \sum_{k} x[k]\,h[n-k] \]

With \( x \) of \( N \) samples and \( h \) of \( M \), the output has \( N + M - 1 \) samples, which is the discrete version of the rule for durations. The computation requires on the order of \( N\cdot M \) multiplications, and there lies the underlying reason the FFT exists: going to frequency, taking a point-by-point product and coming back costs on the order of \( N\log N \), which for large blocks is far less.

It is the same computation as multiplying by hand

Multiplying two polynomials is convolving their coefficients. And multiplying two integers the way it is done in elementary school —product by product, shifting one place and adding the columns— is exactly the same algorithm. Discrete convolution is an operation that everyone has been doing since they learned to multiply.

05In the lab

Exercise 1 · The triangle, by hand

Convolve two rectangular pulses of amplitude 1 and width 2, setting up the pieces separately. Check that the result is a triangle of base 4 and height 2, and that the rules for duration and area hold.

Exercise 2 · The broadening, measured

Inject pulses of 10, 100 and 1000 μs into an RC with \( \tau = 100 \) μs. Measure the width of the output at half height and compare with the rule \( T_y = T_x + T_h \).

Exercise 3 · Discrete convolution, programmed

Program the convolution sum for two 8-sample sequences and check the result against the product of the corresponding polynomials. Count the multiplications that had to be done.

06Common mistakes

  • Forgetting to reflect. It is \( h(t-\tau) \), not \( h(\tau - t) \): the minus sign in front of \( \tau \) is the reflection.
  • Integrating in \( t \) instead of \( \tau \). The variable of integration is \( \tau \); \( t \) is a fixed parameter in each step.
  • Getting the limits wrong. You have to look at where the two signals overlap, and that changes from one piece to the next.
  • Confusing convolution with a product. The point-by-point product is another operation, and it is precisely what convolution is not.
  • Expecting the output to last as long as the input. It lasts the sum of the two durations.

07Self-assessment

What are the four steps of the graphical method?

Reflect \( h \), shift it to \( t \), multiply by \( x \) and measure the area of the product.

How long is the convolution of a 3 ms pulse with a 5 ms one?

8 ms: the durations add.

What does \( x(t) * \delta(t - 2) \) give?

\( x(t-2) \). Convolving with a shifted impulse only shifts.

What does \( h_1 * h_2 \) represent?

The impulse response of the two stages in cascade.

How many samples does the convolution of two sequences of 8 and 5 have?

12, which is \( 8 + 5 - 1 \).

Why is it preferred to work in frequency?

Because convolution becomes a product, and putting stages in cascade becomes multiplying.

08Further reading

  • Alan V. Oppenheim and Alan S. Willsky. Signals and Systems. 2nd ed. (Spanish edition, “Señales y sistemas”), Prentice Hall, 1998. All of chapter 2, with convolution developed graphically piece by piece.
  • Ronald N. Bracewell. The Fourier Transform and Its Applications. 3rd ed., McGraw-Hill, 1999. Convolution treated with a graphical intuition that is hard to find elsewhere.
  • John G. Proakis and Dimitris G. Manolakis. Digital Signal Processing. 4th ed. (Spanish edition, “Tratamiento digital de señales”), Pearson, 2007. Discrete convolution, its computational cost and block-based techniques.
Development of the topic “Convolution” of Signals and Systems (Level 2), based on the curriculum of the UTN Electronic Engineering program, 2023 curriculum — Ordinance No. 1849 of the UTN Higher Council. Back to the Topic Map · catto.ar