I'm asked to find the amplitude response of an LTI system described by the difference equation:

$$ y[n] - ay[n-1] = x[n] $$

After taking the Fourier transform of both sides and some math, I get the frequency response:

$$ H(e^{jw}) = 1/(1-ae^{-jw}) $$

But how do I get the amplitude response from this? I know the amplitude response is

$$ |H(e^{jw})| $$

but I'm unsure how to "get the length" of my frequency response. I have a feeling this is a simple math problem that I'm misunderstanding.

(Disclaimer: this is a question on a past final that I'm using to study. I have the answer to the question (and will post if proof is needed), but I want to understand the method)


3 Answers 3


Although you are in frequency domain you still should be able to get all parameters as you were in time domain. They are different domains but they both should represent the same thing. Time domain represent things in terms of amplitude in respect to time. Frequency domain represent things in terms of amplitude AND PHASE in respect to frequency values. Note that you should have both amplitude and phase in frequency domain, since in the time domain the phase can be represented in the same plot by a shift.

One way to represent these things in frequency domain is by dealing with complex numbers. Complex numbers can be viewed as vectors in a 2D space which have a length (as you said) and an angle. The length represents the output/input ratio and the angle represent the phase shift in comparison also to the input.

So, answering your question, you should calculate the H length to find your output/input ratio. To help you, imagine that:


In other words, its a complex number with always length of 1 and angle \$w\$

You can solve this by two methods:

-Vector method:

imagine that number 1 is \$Z=1+0.i\$ which is a vector to the right, with length 1 and angle \$0\$.

Imagine that \$e^{jw}\$ is a vector that I showed right above

Now add them. Then divive vectors 1 by the vector that you've found.

-Cartesian Coordinates:

represent all in terms of \$Z=a+jb\$ and also \$e^{jw}=cos(w)+jsin(w)\$

and imagine that you have:

\$\large Z = \frac{Z_1}{Z_2+Z_3}\$

and then find length of Z by:

\$|Z| = \sqrt{a^2+b^2}\$


If I remember correctly, during exams we only used to combine templates to draw the (approximate) answer graphically. Stuff like "this term means the amplitude starts falling by -XX dB per octave at frequency YY". Maybe that was because it's not easy/helpful to calculate it symbolicaly.

But it's not hard to calculate a single value of that function. You only need to know complex numbers. I think you even can do it graphically, just keep Euler's formula in mind. Adding two numbers is just adding two vectors, multiplying means you add the angles (phase) and multiply the vector lengths (magnitude).

  • \$\begingroup\$ What you are saying is useful to Bode plots and stuff. You use template forms and add them because the plots are in log scale, so products become sums and divisions become subtracions. Although you use this do draw the final response, and I think he wants to find a function of absolute value of H in terms of w, which should be algebric \$\endgroup\$ Dec 16, 2013 at 21:27

but I'm unsure how to "get the length" of my frequency response.

The DTFT is continuous and periodic.

For the amplitude response, the "length" (in frequency) is \$0 \le \omega \lt \frac{\pi}{T}\$ or, using normalized frequency, \$0 \le \Omega \lt \pi\$

On the other hand, if you're asking how to find the amplitude, recall that:

$$|Z| = \sqrt{ZZ^*}$$


$$|H(e^{j\omega})| = \sqrt{\dfrac{1}{1 - ae^{-j\omega}}\dfrac{1}{1 - ae^{j\omega}}} $$

  • \$\begingroup\$ I think he meant length = absolute value of H \$\endgroup\$ Dec 16, 2013 at 21:25

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