A sinusoid has just been fed into a system.

The system in question has a frequency response given by H(w) = 1/(1+0004jw), where w represents frequency in rads/second.

The input sinusoid has an amplitude of 3, phase of 0 radians and a frequency of 33Hz.

How would I determine the amplitude and phase (between -pi and pi rads) of the sinusoidal output?

Many thanks.

(Keep in mind that an LTI system never changes the frequency of a sinusoid.)

[Paul]

Do you know the theoretical answer to the question and are just asking how to implement that theory in Matlab?

[Jonathan George]

The latter, Paul.

[Sam Chak]

Hi @Jonathan George

Can you do something with this bode() function?

[mag, phase, wout] = bode(sys, w);

https://www.mathworks.com/help/control/ref/lti.bode.html

Think you need to convert the frequency in 33 Hz to rad/s. Do you know how to do it?

[Jonathan George]

I'm unsure, Sam.

[Star Strider]

I don’t understand the leading zeros here:

H(w) = 1/(1+0004jw)

Is there a decimal point missing?

[Jonathan George]

Yes there is, my apologies.

It is H(w) = 1/(1+0.004jw).

[Star Strider]

No worries!

This might be solved easily using Phasor notation.

[Sam Chak]

No wonder, I almost use

s = tf('s');
sys = 1/(1 + 4*s);
bode(sys)

Hz is equivalent to rad/s.

The Bode plot can tell you whether a sinusoidal input signal in amplified or attenuated (in dB), or how much it is shifted in phase when passing through a linear dynamical system at a certain frequency. These info are generally useful when designing low-pass and high-pass filters.

Answer 1

[Star Strider]

Using phasors —

syms omega 
sympref('AbbreviateOutput',false);                                      % Optional
omega = 2*pi*33;                                                        % Radian Frequency
H = 1/(1+0.004*j*omega)                                                 % System

H = 0.5925 - 0.4914i

phasorH = [abs(H), angle(H)]                                            % Phasor = [Amplitude PhaseAngle]

phasorH = 1×2

0.7697 -0.6924

phasorInput = [3, 0]

phasorInput = 1×2

3 0

phasorOutput = [phasorH(1)*phasorInput(1), phasorH(2)+phasorInput(2)]   % Multiply Amplitudes, Add Phases

phasorOutput = 1×2

2.3091 -0.6924

So, the output has an amplitude of 2.3091 and a phase angle of -0.6924 radians.

I haven’t routinely worked with phasors in a while (since grad school, back in the Precambrian) so check this. However, I believe the approach is correct.

.

[Jonathan George]

I have confirmed it is correct! Thanks again, Star :)

[Star Strider]

As always, my pleasure!

Answer 2

[Sam Chak]

Hi @Jonathan George

This is sort of the "control theorist's way", if you are interested to learn.

s = tf('s');
sys = 1/(1 + 0.004*s)
omega = 33*2*pi;
[mag, phase, wout] = bode(sys, omega)
Input_Amplitude = 3
Output_Amplitude = Input_Amplitude*mag
Output_phase_in_radian = (pi/180)*phase     % the unit degree is commonly used

sys =

1

-----------

0.004 s + 1

Continuous-time transfer function.

mag =

0.7697

phase =

-39.6716

wout =

207.3451

Input_Amplitude =

3

Output_Amplitude =

2.3091

Output_phase_in_radian =

-0.6924