# Complex signal transform

I must strengthen my knowledge in the mathematical properties of signals, and would like to know the following thing:

• if I have a complex signal, written in magnitude and phase, it's like if I have 1 signal that holds 2 signal himself? I have this question because I have to do the inverse transformation of a signal in the frequency domain , that is complex and I have the magnitude and phase plot. Is it right that my phase, when I do the inverse- transform retards my signal ( the inverse transform of the magnitude) in the time domain? Thank you so much.. I have not on focus how the phase works yet

Thank you for the answer, I know what you have written... my question is: If I have one complex signal in the magnitude and phase form, do I have to consider magnitude and phase as two different signals? Look The signal is written in magnitude with a complex exponential with the phase... when I do the inverse transform the exponential is in the time domain a delay ... and this it feels strange that one signal bring with him a delay ... it's like if phase is another signal , that retards the inverse transform of the magnitude. I hope I have explained my question in the right way

• You need to convert the amplitude and phase to a complex signal by calculating the real components as AmplitudeXcos(phase) and the imaginary part being AmplitudeXsin(phase). The FFT (or inverse) is carried out on complex vectors. – Moti Jan 13 '16 at 23:17
• You seem to have two questions. I have answered the question about a complex signal "holding" two signals, but I don't understand your question about the phase and the inverse transform. If you clarify, I may be able to answer that question too. – MBaz Jan 13 '16 at 23:18
• I've expanded my answer; please take a look. – MBaz Jan 14 '16 at 15:16

A complex signal can be written as $s(t)=a(t)+jb(t)$, which indeed corresponds to two real signals ($a(t)$ and $b(t)$) "embedded" in the single signal $s(t)$. What makes this interesting is that $a(t)$ and $b(t)$ can be easily recovered from $s(t)$. This is true whether you write the complex signal in rectangular or polar coordinates.

Now, regarding how to interpret the phase of a signal. Let's say that the Fourier transform of $s(t)$ is $S(f)=|S(f)|e^{j\phi(f)}$. Imagine that you have an infinitely narrow band-pass filter with center frequency $f_0$. Then, you connect the signal $s(t)$ to the filter's input and see the output in an oscilloscope. You'll see a sinusoidal wave, with frequency $f_0$, amplitude $|S(f_0)|$ and phase $\phi(f_0)$. Here, "phase" indicates the difference, in radians, between a zero-phase cosine $\cos(2\pi f_0t)$ and the sinusoid you see in the scope.

In that sense, the phase $\phi(f_0)$ can be thought of as a "delay"; the sinusoid in the scope is just a cosine with a delay, $\cos(2\pi f_0t+\phi(f_0))$.

As an application, consider a signal that goes through an LTI system. The system doesn't touch $|S(f)|$ but it modifies the phase $\phi(t)$ -- that is, it changes the phase of each individual sinusoid. If this results in the same delay for each and every sinusoid, then the system output will be a delayed version of $s(t)$. If some of the sinusoids suffer different delays, then the signal at the system's output will be a distorted version of $s(t)$.

• I have edited my answer... can you look? – Alberto Jan 14 '16 at 13:25
• Thank you very much... you have been really clearly. Maybe, since I am at the first course of signals, I have not all the concept, and I have never seen an oscilloscope. so, when you say that in the filter's output I find a sinusoidal wave with a delay given by the phase in the narrow filter, right? And I can see the phase of a filter ( function of frequency ) , as a delay for every sinusoidal wave of the fourier series at a every frequency, right? I've not understand in my exercise, why my signal 's phase has an effect of delay in my signal x(t) ... in my case x(t-t0) cause phase was -2 pi t0 – Alberto Jan 14 '16 at 18:42
• @Alberto I used an oscilloscope to try to give intuition for the result... maybe you can get another student or professor to show you one? In any case, yes, you'll have a sine wave at the filter's output. Also, as I mentioned, if the filter has linear phase, then the output is indeed a delayed (and filtered) version of the input. If the phase is not linear, then the output is also distorted because different signal components suffer different delays. – MBaz Jan 15 '16 at 23:52
• And which kind of signal I see at the output filter with phase not linear? Is it no more a sinusoidal wave? – Alberto Jan 16 '16 at 13:37

You can write the overall frequency response in the format:

$$\displaystyle X(f) = A(f) e^{j\phi(f)},$$

where $\displaystyle A(f) = |X(f)|$ is the magnitude response and $\displaystyle \phi (f) = \angle X(f)$ is the phase response. You can obtain both through the graphics you have shown in your question (let me know if you have problems about this).

Next you just need to compute the inverse transform.

• I knew this... maybe I have not explained in the right way :) – Alberto Jan 14 '16 at 18:42