# bandwidth between the signal and the channel

I have a signal with a bandwidth $$B=4$$ kHz and the bandwitdh is between $$6$$ kHz and $$10$$ kHz. I want to transmit this signal in a channel with bandwidth $$B'=4$$ kHz but for the channel the bandwidth is between $$1$$ kHz and $$5$$ kHz.

This is my question: Can we pass a signal in this channel even if the bandwidth support is not the same?

• The short answer is no, you can't. But you mention a very specific signal bandwidth (4 kHz) and then a range (6 to 10 kHz). What is the actual signal bandwidth? – MBaz Mar 11 at 21:02

If I understand correctly you would like to pass a bandpass signal of 4Khz bandwidth in the frequency range 6 to 10 Khz, but you would like to transmit this directly over a channel which has good channel gain in 1 to 5Khz. Clearly, the signal when passed over a channel with this frequency response will suffer complete degradation.

Hence, we cannot transmit this signal directly over this channel. You could first downcovert the signal by 5 Khz and then transmit.

Yes you can shift its spectrum to fit into the band of $$[1k, 5k]$$ Hz.

The following will do: $$y(t) = \cos(2\pi ~5000~ t) \cdot x(t)$$

A new band of will appear at $$[11k, 15k]$$ Hz, that must either be filtered out by you before adding to the channel, or the channel itself will do it for you.

The answer to your direct question is no because the channel will not pass the frequencies where the signal is.

To put in some practical sense, it will happen in the high frequency (HF = 3-30 MHz) regime where many factors affect whether or not certain frequencies propagate well. For example, a signal may propagate at a given frequency during the certain hours of the day, specific weather conditions, and even seasons.

This doesn't stop you from shifting your signal to a different frequency band though. For example, if your signal was shifted down to match the channel then you could transmit with (hopefully) only a small amount of attenuation given that you do a good job matching your signal bandwidth to the channel. The mathematical operation is laid out in @Fat32's answer.