why bandwidth returns an "inf"?

I have this tf with the following bode plot

k = 1
z = [-1 4 0]
p = [-10+4*j -10-4*j -3 -3 -3]
sys = zpk(z,p,k);
display(sys);

figure(1);
bode(sys);
grid

Why, in this case, the bandwidth(sys) retruns an "Inf"?

And is it only a problem in matlab to calculate the bandwidth or a bandpass exist but I have to calculate in another way?

it is not clear to me looking the plot.

thanks in advance.

Answer 1

[Star Strider]

It returns Inf because the bandwidth function requires a monotonically-decreasing signal, not one with a passband.

Try this —

k = 1

k = 1

z = [-1 4 0]

z = 1×3

-1 4 0

p = [-10+4*j -10-4*j -3 -3 -3]

p = 
 -10.0000 + 4.0000i -10.0000 - 4.0000i  -3.0000 + 0.0000i  -3.0000 + 0.0000i  -3.0000 + 0.0000i

sys = zpk(z,p,k);
display(sys);

sys =
 
        s (s+1) (s-4)
  -------------------------
  (s+3)^3 (s^2 + 20s + 116)
 
Continuous-time zero/pole/gain model.

[mag,phs,wout] = bode(sys);
smag = squeeze(mag)

smag = 61×1

1.0e+00 * 0.0001 0.0001 0.0002 0.0002 0.0002 0.0003 0.0003 0.0004 0.0004 0.0005

sphs = squeeze(phs)

sphs = 61×1

267.5632 267.3283 266.8726 266.3356 265.7005 264.9465 264.0464 262.9653 261.6568 260.0597

dB3 = 10^(-3/20)

dB3 = 0.7079

[maxmag,idx] = max(smag)

maxmag = 0.0061

idx = 28

wascdB3 = interp1(smag(1:idx), wout(1:idx), dB3*maxmag)

wascdB3 = 2.4349

wdscdB3 = interp1(smag(idx:end), wout(idx:end), dB3*maxmag)

wdscdB3 = 11.1138

bndwdh = wdscdB3 - wascdB3

bndwdh = 8.6789

figure(1);
semilogx(wout, mag2db(smag))
hold on
plot([wascdB3, wdscdB3], [1 1]*mag2db(maxmag*dB3), '+-r')
hold off
grid
xlabel('Frequency (rad/s)')
ylabel('Magnitude(dB)')
text(mean([wascdB3, wdscdB3]), mag2db(maxmag*dB3), sprintf('\\uparrow\nBandwidth = %.2f',bndwdh), 'Horiz','center', 'Vert','top')

The red line is between the -3 dB points on the passband.

.

Answer 2

[Chunru]

fb = bandwidth(sys) returns the bandwidth of the SISO dynamic system model sys. The bandwidth is the first frequency where the gain drops below 70.79% (-3 dB) of its DC value. The bandwidth is expressed in rad/TimeUnit, where TimeUnit is the TimeUnit property of sys.

So this bandwidth is applicable for a low pass filter only.

For your system, DC response is 0, so the bandwidth is Inf

[Antonio Del Vecchio]

Ok, and this is now clear to me. Thnaks a lot.

So, THIS command "bandwith" acts in this way and, then, cannot work in this case.

But, looking at the plot, do you agree that there is a "bandpass point"? If yes, do we have a command to find it or I have to do it "manually" (and in which way)?

[Chunru]

You can write a code to do that:

• Find the peak response
• Find the first points to the left of peak that is 3dB lower
• Find the first points to the right of peak that is 3dB lower
• Then compute the bandwidth.

Note that this is not full-proof bandwidth for arbitrary response.