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Question sheet 2: Solution
A wireless transmitter sends data with a symbol period T using a transmit filter with an impulse response hT(t) which is a rectangular pulse with length T.
Referring to slide 3:8, what should be the impulse response of the receive filter hR(t) to maximise received signal to noise ratio?
The receive filter hR(t) should of course also be a rectangular filter of duration T. Then we can write:
The combined impulse response of the transmit and receive filter, h() as shown on slide 4:9 is given by the convolution of the transmit and receive filter impulse responses:
Find and sketch h() in this case.
Combined response:
i.e. the result is a triangular impulse response, with peak T – however we normalise this to 1 in subsequent calculations.
The following are the magnitudes, delays and phases of multipath components on a wireless channel. Use the first formula on slide 4:10 to calculate the tap weights gj of a symbol-spaced tapped delay line representing this channel, if the transmit and receive filters discussed above are used.
i/T |hi| |hi|2 i hi
0 1 1 0 1
0.25 0.7 0.49 π/4 0.495 + 0.495 j
0.5 0.6 0.36 π/2 0.6 j
0.75 0.5 0.25 π -0.5
1.25 0.5 0.25 - π/2 -0.5 j
1.5 0.2 0.04 0 0.2
2 0.2 0.04 π/2 0.2 j
i/T hi Tap 1:
Tap 1:
Tap 2:
Tap 2:
0 1 1 1 0 0
0.25 0.495 + 0.495 j 0.75 0.371+ 0.371 j 0.25 0.124 + 0.124 j
0.5 0.6 j 0.5 0.3 j 0.5 0.3 j
0.75 -0.5 0.25 -0.125 0.75 -0.375
1.25 -0.5 j 0 0 0.75 - 0.375 j
1.5 0.2 0 0 0.5 0.1
2 0.2 j 0 0 0 0
Sum 1.246 + 0.671 j -0.151 + 0.049 j
Also sketch on a second graph , , , etc. Sketch on the same graph the squared magnitude of the channel impulse response. Assuming now that the phases of the multipaths are random (but delays and magnitudes are as listed above), use the second formula on slide 4:10 to find the root mean square magnitudes of the tap weights, , for the channel model.
We now assume that the exact magnitudes and phases of the multipaths are random, with root mean square magnitude as given in the table above. Now we calculate the mean square tap weights using:
i/T |hi| |hi|2 Tap 1:
Tap 1:
Tap 2:
Tap 2:
Tap 3:
Tap 3:
0 1 1 1 1
0.25 0.7 0.49 0.5625 0.276
0.5 0.6 0.36 0.25 0.09
0.75 0.5 0.25 0.0625 0.0156
1.25 0.5 0.25 0 0
1.5 0.2 0.04 0 0
2 0.2 0.04 0 0
Sum 1.381
Sqrt 1.175
Question sheet 2: Solution
A wireless transmitter sends data with a symbol period T using a transmit filter with an impulse response hT(t) which is a rectangular pulse with length T.
Referring to slide 3:8, what should be the impulse response of the receive filter hR(t) to maximise received signal to noise ratio?
The receive filter hR(t) should of course also be a rectangular filter of duration T. Then we can write:
The combined impulse response of the transmit and receive filter, h() as shown on slide 4:9 is given by the convolution of the transmit and receive filter impulse responses:
Find and sketch h() in this case.
Combined response:
i.e. the result is a triangular impulse response, with peak T – however we normalise this to 1 in subsequent calculations.
The following are the magnitudes, delays and phases of multipath components on a wireless channel. Use the first formula on slide 4:10 to calculate the tap weights gj of a symbol-spaced tapped delay line representing this channel, if the transmit and receive filters discussed above are used.
i/T |hi| |hi|2 i hi
0 1 1 0 1
0.25 0.7 0.49 π/4 0.495 + 0.495 j
0.5 0.6 0.36 π/2 0.6 j
0.75 0.5 0.25 π -0.5
1.25 0.5 0.25 - π/2 -0.5 j
1.5 0.2 0.04 0 0.2
2 0.2 0.04 π/2 0.2 j
i/T hi Tap 1:
Tap 1:
Tap 2:
Tap 2:
0 1 1 1 0 0
0.25 0.495 + 0.495 j 0.75 0.371+ 0.371 j 0.25 0.124 + 0.124 j
0.5 0.6 j 0.5 0.3 j 0.5 0.3 j
0.75 -0.5 0.25 -0.125 0.75 -0.375
1.25 -0.5 j 0 0 0.75 - 0.375 j
1.5 0.2 0 0 0.5 0.1
2 0.2 j 0 0 0 0
Sum 1.246 + 0.671 j -0.151 + 0.049 j
Also sketch on a second graph , , , etc. Sketch on the same graph the squared magnitude of the channel impulse response. Assuming now that the phases of the multipaths are random (but delays and magnitudes are as listed above), use the second formula on slide 4:10 to find the root mean square magnitudes of the tap weights, , for the channel model.
We now assume that the exact magnitudes and phases of the multipaths are random, with root mean square magnitude as given in the table above. Now we calculate the mean square tap weights using:
i/T |hi| |hi|2 Tap 1:
Tap 1:
Tap 2:
Tap 2:
Tap 3:
Tap 3:
0 1 1 1 1
0.25 0.7 0.49 0.5625 0.276
0.5 0.6 0.36 0.25 0.09
0.75 0.5 0.25 0.0625 0.0156
1.25 0.5 0.25 0 0
1.5 0.2 0.04 0 0
2 0.2 0.04 0 0
Sum 1.381
Sqrt 1.175