# dsp, determine the time signal x(t) ?

#### MrSalts

Joined Apr 2, 2020
2,620
good morning everyone, in the picture below we have a spectrum of a signal
View attachment 261928
The question is to determine the time signal x(t) ?
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#### alitronic

Joined Jun 13, 2020
37
I found that x(t)=1+3cos(2*pi*500*t)+2cos(2*pi*1000*t)
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#### alitronic

Joined Jun 13, 2020
37
I found that x(t)=1+3cos(2*pi*500*t)+2cos(2*pi*1000*t)
but what the expression of the signal becomes if it is filtered by a low pass filter with a frequency of 700 Hz.

#### MrAl

Joined Jun 17, 2014
9,633
I found that x(t)=1+3cos(2*pi*500*t)+2cos(2*pi*1000*t)
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?

• alitronic

#### alitronic

Joined Jun 13, 2020
37
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?
Good evening, the expression I write is of the original signal (before the filter)

#### MrAl

Joined Jun 17, 2014
9,633
Good evening, the expression I write is of the original signal (before the filter)
Do you mean that after the filter you get that x response?
Because the x response does not look like that expression.

#### alitronic

Joined Jun 13, 2020
37
Do you mean that after the filter you get that x response?
Because the x response does not look like that expression.
no, I mean x(t) is the expression before the ideal low-pass filter of 700 Hz.

#### MrAl

Joined Jun 17, 2014
9,633
no, I mean x(t) is the expression before the ideal low-pass filter of 700 Hz.
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.

#### alitronic

Joined Jun 13, 2020
37
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, 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

#### MrAl

Joined Jun 17, 2014
9,633
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
Hello 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.

• alitronic

#### alitronic

Joined Jun 13, 2020
37
Hello 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.
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).
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?

#### MrAl

Joined Jun 17, 2014
9,633
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).
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?
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.

• alitronic

#### alitronic

Joined Jun 13, 2020
37
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.
Thank you very much.