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DT transfer function; drawing a Bode plot

Short answer

Impulse response h[n] is the output for δ[n] input. Transfer function H(z)=Y(z)/X(z) equals Z-transform of h[n]. Bode plot procedure: factor H(s) into standard form, identify corner frequencies, draw asymptotic magnitude (dB) and phase (degrees) plots by summing contributions, then correct.

Define transfer function and impulse response in discrete-time. Briefly explain the procedure of drawing a bode plot.

Impulse response (discrete time)

The impulse response h[n]h[n] of a discrete-time LTI system is the output when the input is the unit impulse δ[n]\delta[n] and the system is initially at rest:

δ[n]→h[n]\delta[n] \rightarrow h[n]

Because any input can be written as x[n]=∑kx[k]δ[n−k]x[n]=\sum_k x[k]\delta[n-k], the output of an LTI system is fully fixed by h[n]h[n] through the convolution sum y[n]=x[n]∗h[n]y[n]=x[n]*h[n].

Transfer function (discrete time)

The transfer function is the ratio of the transform of the output to the transform of the input (zero initial conditions). It equals the transform of h[n]h[n]:

H(z)=Y(z)X(z)=∑n=−∞∞h[n]z−n,H(ejω)=Y(ejω)X(ejω)=∑nh[n]e−jωnH(z)=\frac{Y(z)}{X(z)}=\sum_{n=-\infty}^{\infty}h[n]z^{-n}, \qquad H(e^{j\omega})=\frac{Y(e^{j\omega})}{X(e^{j\omega})}=\sum_{n}h[n]e^{-j\omega n}

H(ejω)H(e^{j\omega}) (the frequency response) is H(z)H(z) evaluated on the unit circle z=ejωz=e^{j\omega}. Example: for y[n]=0.5y[n−1]+x[n]y[n]=0.5y[n-1]+x[n], H(ejω)=11−0.5e−jωH(e^{j\omega})=\frac{1}{1-0.5e^{-j\omega}} and h[n]=(0.5)nu[n]h[n]=(0.5)^n u[n].

Procedure for drawing a Bode plot

A Bode plot is a pair of graphs: magnitude in dB, 20log⁡10∣H(jω)∣20\log_{10}|H(j\omega)|, and phase ∠H(jω)\angle H(j\omega), both against log⁡10ω\log_{10}\omega.

  1. Write H(jω)H(j\omega) in standard (time-constant) form, e.g. H(jω)=K(1+jω/z1)jω(1+jω/p1)H(j\omega)=\frac{K(1+j\omega/z_1)}{j\omega(1+j\omega/p_1)}.
  2. Identify the factors: constant KK, poles/zeros at the origin, simple real poles/zeros, and quadratic factors. Note each corner frequency (ω=z1,p1,…\omega=z_1, p_1,\dots).
  3. Draw the asymptote of each factor:
    • constant: flat line at 20log⁡K20\log K dB, phase 0° (or 180° if K<0K<0);
    • pole at origin: line of −20-20 dB/decade through 0 dB at ω=1\omega=1, phase −90∘-90^\circ;
    • simple pole: 0 dB up to the corner, then −20-20 dB/decade; phase goes from 0° to −90∘-90^\circ (−45° at the corner);
    • simple zero: same but +20+20 dB/decade and +90∘+90^\circ;
    • quadratic pole: −40-40 dB/decade after ωn\omega_n, phase 0° to −180°.
  4. Add the magnitude asymptotes (dB adds because log of a product is a sum), starting from the lowest frequency; the slope changes by ±20 dB/dec at each corner.
  5. Add the phase curves of all factors at several frequencies.
  6. Correct the asymptotic plot if needed (about 3 dB at a simple corner; resonance peak for low damping).

For a discrete-time system the same idea is used with H(ejω)H(e^{j\omega}) plotted for 0≤ω≤π0\le\omega\le\pi (it is periodic in 2π2\pi), usually on a linear frequency axis.

 dB
  |------\            magnitude: flat, then
  |       \  -20 dB/dec   rolls off after corner
  |        \
  +---------+--------> log w
          w_c

✓ Answer checked by Claude Opus 5.5 (high effort)

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