Show us how you approached the problem and why you are having troubles. If you simply need answers, take a subscription to Chugg.comgood morning everyone, in the picture below we have a spectrum of a signal
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The question is to determine the time signal x(t) ?
Please
Show us how you approached the problem and why you are having troubles. If you simply need answers, take a subscription to Chugg.com
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but what the expression of the signal becomes if it is filtered by a low pass filter with a frequency of 700 Hz.I found that x(t)=1+3cos(2*pi*500*t)+2cos(2*pi*1000*t)
Hello,I found that x(t)=1+3cos(2*pi*500*t)+2cos(2*pi*1000*t)
Good evening, the expression I write is of the original signal (before the filter)Hello,
What did you use to get that expression?
Also, the simplest way to find out the response with a low pass filter would be to create a drive signal by taking the Laplace Transform of your signal, then apply it to a filter i presume has a gain of 1 for DC and 1/sqrt(2) at 700Hz. You can use an RC filter then multiply the output by the factor that gives you a gain of 1 for DC.
Does this make sense to you?
Do you mean that after the filter you get that x response?Good evening, the expression I write is of the original signal (before the filter)
no, I mean x(t) is the expression before the ideal low-pass filter of 700 Hz.Do you mean that after the filter you get that x response?
Because the x response does not look like that expression.
Hello again,no, I mean x(t) is the expression before the ideal low-pass filter of 700 Hz.
Hello, thank you for.Hello again,
Oh ok, well i asked because i got something different and it checks out in a simulation. So i have to suggest that you either go over this again or explain in detail mathematically how you got your result so we can look it over.
Check out the Fourier components for your purposed solution.
Hello again,Hello, thank you for.
Please correct me if I am wrong.
In the billateral spectrum, there is a Fourier components at 0 Hz, I consider it as the average value (mean) of the function, hence the term 1 in the expression.
There are also two symetric terms at 500 Hz and -500 Hz with amplitude 1,5 who in the time domain becom 3 cos (2*pi*500)
and finally two symetric terms at 1000 Hz and -1000 Hz with amplitude 1 who become in time domain 2 cos (2*pi*1000)
Hence, the inal expression is x(t)=1+3cos(2*pi*500*t)+2cos(2*pi*1000*t)
Thanks a lot again
Thank you, correct me again is I am wrongHello again,
Yes you are absolutely correct. For some reason i was looking at the one sided spectrum.
It's more common to use the one sided spectrum but as you note there are times when we cant do that.
It's also easy to look at the one sided in a simulator.
Hello again,Thank you, correct me again is I am wrong
The remaining questions of the exercise are
1- Determine the expression of the signal at the output of an ideal low-pass filter (with fc=700 Hz).
My answer was 1+3cos(2*pi*500Hz)
2- The filtered signal is applied to the input of a sampler who has a sampling period Ts=0.8 ms. What is the expression at the ouput of the sampler?
NO ANSWER.
Help please.
Thank you very much.Hello again,
For #1, since an ideal low pass filter passes amplitudes under the cutoff frequency perfectly and cuts everything else to zero, that would mean that it would pass all frequencies from 0 to 700Hz exactly as they are input and cut every other frequency to zero.
So yes that appears to be correct.
For this question though, it would be very interesting for you to calculate the amplitude and phase responses of the output when using a non ideal LP filter, such as a first order RC filter, then a second order RC filter. That's not too hard to do and a bit more interesting and informative. For the second order filter you could make both resistors the same and both capacitors the same, then try making the second set different than the first set.
For #2, what have you done in the past that might be related to this question?
Are you looking for a Dirac type function or a Z transform or something else?
In the first case i guess you would convolve the input with the Dirac comb function
In the second case you would just calculate the Z transform.
I am sure there is a lot written on both those on the web too.