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First year student so please excuse my lack of knowledge.

As i understand i need to use convolution which is: $$ y^{out}(t) = u^{in}(t) * h(t) $$ Or maybe my thoughts are wrong so please correct me.

But still i do not understand how to draw magnitude spectrum of output signal.

I did find images of convolution but still do not understand.

Task

Edit:

So, using Dan Boschen hint i tried to solve this task. I assumed than i need to multiply two graphs as they were in frequency domain.

try

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  • $\begingroup$ What class are you taking? Did you already cover the Fourier Transform and spectra ? $\endgroup$
    – Hilmar
    Commented Jun 24, 2023 at 20:32
  • $\begingroup$ I am taking signal theory class, yes we did cover Fourier Transform as well as Phase and Magnitude spectrum $\endgroup$ Commented Jun 24, 2023 at 20:40
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    $\begingroup$ Just something additonal: Get comfortable with writing the units for the frequency axis because it becomes really important once you study Sampling (although this book hasn't written the units since it's not important for this exercise). In this case your FT is the CTFT (Continous-Time Fourier Transform) of a time-domain signal so the units are rad/s or radian per second. $\endgroup$ Commented Jun 24, 2023 at 21:43

1 Answer 1

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As a hint, the key points to understand are convolution in the time domain is a product in the frequency domain. The “spectrum” is the frequency domain; the result of the Fourier Transform of the time domain signal. It will be complex and therefore have a magnitude and phase. The magnitude spectrum would be the absolute value of the complex values, and similarly if needed (not in this question) the phase spectrum would be the phase of the complex values.

Given the magnitude spectrums of the input and of the system are already provided, determining the magnitude of the output spectrum should be very straight forward.

Show your work by updating your question with what you think the answer should be and why, and I will be happy to check that.

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  • $\begingroup$ So, i updated my question with a photo of my work, please tell me if this is what you meant. $\endgroup$ Commented Jun 24, 2023 at 21:20
  • $\begingroup$ Yes it looks like you understood well. I agree with your answer. Nice job $\endgroup$ Commented Jun 24, 2023 at 21:22
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    $\begingroup$ Thank you for your help! $\endgroup$ Commented Jun 24, 2023 at 21:24

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