Sketch the spectrum of the DSB-SC signal

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Fatima Hasan
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Homework Statement
Attached below.
Relevant Equations
cos(t) <--> 1/2 [δ (f-fc) + δ(f+fc) ]
Here's my work:

5k%29%20+%5Cdelta%20%28f-5.5k%29+%20%5Cdelta%28f+5.gif


1615761629365.png


Could someone please confirm my answer?
 
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Fatima Hasan said:
Could someone please confirm my answer?

I think the frequency spectrum is correct. However, I think there is perhaps a little typo in the working - it doesn't change the final answer though.

Fatima Hasan said:

I think the line ## S(f) = \frac{1}{2} ( M(f - 500) + M(f + 500) ) ## should instead have ## 5000 ## in the brackets rather than 500 right? We are convolving ## M(f) ## with ## \frac{1}{2}(\delta(f - 5000) + \delta(f + 5000)) ## which would lead to:

[tex]S(f) = \frac{1}{2} ( M(f - 5000) + M(f + 5000) )[/tex]
Then we substitute ## M(f) = \frac{1}{2}(\delta(f - 500) + \delta(f + 500)) ## for the first part which should lead to the same final expression that you have.
 
Fatima Hasan said:
Homework Statement:: Attached below.
Relevant Equations:: cos(t) <--> 1/2 [δ (f-fc) + δ(f+fc) ]

Here's my work:

View attachment 279784

View attachment 279782

Could someone please confirm my answer?
Can't read your notes but it looks like your S(f) is the carrier in the frequency spectrum.

"DSB-SC" has no carrier power in the spectrum.
The sidebands are above and below the carrier frequency.
If you multiply a signal ## sin(\omega_st ## by the carrier ## sin(\omega_ct ## you get DSB-SC spectrum as high school trig will reveal. Sidebands at ## (\omega_c + \omega_s) ## and ##\omega_c - \omega_s ##.
 
rude man said:
Can't read your notes but it looks like your S(f) is the carrier in the frequency spectrum.

"DSB-SC" has no carrier power in the spectrum.
The sidebands are above and below the carrier frequency.
If you multiply a signal ## sin(\omega_st ## by the carrier ## sin(\omega_ct ## you get DSB-SC spectrum as high school trig will reveal. Sidebands at ## (\omega_c + \omega_s) ## and ##\omega_c - \omega_s ##.
Sorry, I don't quite understand. From the image in the picture, it looks as if there are only the two side bands present (and the -ve frequency versions as well)...