Bandwidth of a channel and QPSK

I know the meaning of bandwidth of a system or a signal but what do you mean by "bandwidth of the channel" as you see in the Shannon's formula.

In the Shannon-Hartley theorem for channel capacity
C = B log2(1+S/N)
Wikipedia says "B" is the bandwidth of the channel and I understand that it is related to the channel capacity but what is "B" as channel capacity is the maximum information that the channel can carry.

Based on this Bandwidth I got another doubt too.

When I was learning Qpsk I came across this sentence
"For a given bit rate the transmission bandwidth required for Qpsk is half as that of Psk."
I know that in Qpsk a symbol contains two bits and the duration of the symbol is twice that of the PSK but I couldn't quite understand the above sentence.

Thanks in advance.

The bandwidth of a system is the difference between the highest and lowest frequencies which the system can carry.

e.g. If a system can carry frequencies between 200Hz and 4kHz, its bandwidth (the difference between those two frequencies) is 3.8kHz.

A system which can carry frequencies between, say, 10MHz and 100MHz, has a bandwidth of 90MHz.

Does that mean the channel capacity changes with the given input signal to the channel? Does that mean both "bandwidth of the channel" and "bandwidth of the signal/system" are one and the same?

Thanks in advance.

The maximum bandwidth efficiency (R_b/W) of BPSK is 1 b/s/Hz, while the maximum bandwidth efficiency of QPSK is 2 b/s/Hz. Now if you want to transmit data of rate 10 Mbps you will need 10 Mhz for BPSK and 5 MHz for the QPSK.

Please answer to my first question. What do you mean by the term "Bandwidth of a channel" ?. How is it different from the term "Bandwidth of the signal"?

The definition of bandwidth is same for both the signal and the channel. What is the bandwidth of a signal? For example, a voice signal has significant components between 300Hz and 3300 Hz, so the bandwidth of the voice signal is 3 KHz. So, we need a channel with bandwidth at least 3 KHz to transmit the voice signal over it. Channels usually have bandwidth over certain range of frequencies. For example, if we assume that the bandwidth of the channel is 2W, and the central frequency is f, then the lowest frequency component will be f-W and the largest frequency component will be f+W, so we need to transfer the baseband frequency component of the signal to that range by means of carrier.

Okay Thanks. Now answer me this question

"For a given bit rate the transmission bandwidth required for Qpsk is half as that of Psk."

I know that in Qpsk a symbol contains two bits and the duration of the symbol is twice that of the PSK but I couldn't quite understand the above sentence.

Sir, I answered this question in post #4. Anyway here it is in another way:

In pass-band communication, the minimum required bandwidth for a transmission rate of Rs symbol/s is Rs (assuming rectangular pulse shape). In BPSK the symbol rate is the same as the bit rate, i.e.: Rs=Rb. So, the minimum required bandwidth for BPSK is WB=Rb. For QPSK, Rs=Rb/2, so WQ=Rb/2. It is obvious that the bandwidth required by QPSK is half of that required by BPSK.