Sketch the spectrum of the DSB-SC signal

In summary, the conversation discusses the correct frequency spectrum for a given problem and confirms that there may be a small typo in the working, but it does not change the final answer. It also mentions the absence of carrier power in the spectrum for DSB-SC and explains the presence of sidebands above and below the carrier frequency. The conversation ends with a request for clarification on the image provided.
  • #1
Fatima Hasan
319
14
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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  • #2
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.
 
  • #3
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 ##.
 
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  • #4
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)...
 

1. What is DSB-SC signal?

The DSB-SC (Double-Sideband Suppressed Carrier) signal is a type of amplitude modulation (AM) in which the carrier signal is completely removed, leaving only the upper and lower sidebands. This results in a more efficient use of bandwidth compared to traditional AM signals.

2. How is the spectrum of a DSB-SC signal sketched?

The spectrum of a DSB-SC signal is typically sketched by plotting the amplitude (or power) of the signal against frequency. This results in a spectrum with two peaks at the frequencies of the upper and lower sidebands, with no peak at the carrier frequency.

3. What is the bandwidth of a DSB-SC signal?

The bandwidth of a DSB-SC signal is equal to the sum of the bandwidths of the upper and lower sidebands. This is because both sidebands contain the same information, just at different frequencies.

4. How does DSB-SC differ from other types of AM?

Unlike traditional AM signals, DSB-SC does not contain a carrier signal, resulting in a more efficient use of bandwidth. Additionally, DSB-SC signals are more susceptible to noise and interference, as there is no carrier signal to help with signal recovery.

5. What are the applications of DSB-SC signals?

DSB-SC signals are commonly used in radio and television broadcasting, as well as in telecommunications and wireless communication systems. They are also used in radar systems and medical imaging devices.

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