In the past, we had discussed BER for BPSK in flat fading Rayleigh channel. In this post, lets discuss a frequency selective channel with the use of Zero Forcing (ZF) equalization to compensate for the inter symbol interference (ISI). For simplifying the discussion, we will assume that there is no pulse shaping at the transmitter. The ISI channel is assumed to be a fixed 3 tap channel.
Transmit symbol
Let the transmit symbols be modeled as
, where
is the symbol period,
is the symbol to transmit,
is the transmit filter,
is the symbol index and
is the output waveform.
For simplicity, lets assume that the transmit pulse shaping filter is not present, i.e .
So the transmit symbols can be modeled by the discrete time equivalent

Figure: Transmit symbols
Channel Model
Lets us assume the channel to be a 3 tap multipath channel with spacing i.e.
Figure: Channel model (3 tap multipath)
In addition to the multipath channel, the received signal gets corrupted by noise , typically referred to as Additive White Gaussian Noise (AWGN). The values of the noise
follows the Gaussian probability distribution function,
with
mean and
variance .
The received signal is
, where
is the convolution operator.
Zero Forcing Equalization
Objective of Zero Forcing Equalization is to find a set of filter coefficients which can make
.
After equalization
.
Note:
The term causes noise amplification resulting poorer bit error rate performance.
Deriving the equalization coefficients
From the post on toeplitz matrix,we know that convolution operation can be represented as matrix multiplication.
% Matlab code for using Toeplitz matrix for convolution clear all x = [1:3]; h = [4:6]; xM = toeplitz([x zeros(1,length(h)-1) ], [x(1) zeros(1,length(h)-1) ]); y1 = xM*h'; y2 = conv(x,h); diff = y1'-y2 diff = [ 0 0 0 0 0 ]
Using similar matrix algebra and assuming that the coefficients has 3 taps, the equation
can be equivalently represented as,
Solving for , we have
.
If we assume that has 5 taps,
Solving for , we have
.
Example
% Assuming a 3 tap channel as follows ht = [0.2 0.9 0.3]; L = length(ht); kk = 1; hM = toeplitz([ht([2:end]) zeros(1,2*kk+1-L+1)], [ ht([2:-1:1]) zeros(1,2*kk+1-L+1) ]); d = zeros(1,2*kk+1); d(kk+1) = 1; c = [inv(hM)*d.'].';
The frequency response of the channel and the equalizer
are shown below:

Figure: Frequency response of the channel and the equalizer
Simulation Model
Click here to download: Matlab/Octave script for computing BER for BPSK with 3 tap ISI channel with Zero Forcing Equalization
The attached Matlab/Octave simulation script performs the following:
(a) Generation of random binary sequence
(b) BPSK modulation i.e bit 0 represented as -1 and bit 1 represented as +1
(c) Convolving the symbols with a 3-tap fixed fading channel.
(d) Adding White Gaussian Noise
(e) Computing the equalization filter at the receiver – the equalization filter is 3, 5, 7, 9 taps in length
(f) Demodulation and conversion to bits
(g) Counting the number of bit errors
(h) Repeating for multiple values of Eb/No
The simulation results are as shown in the plot below.

Figure: BER plot for BPSK in a 3 tap ISI channel with Zero Forcing equalizer
Observations
1. Increasing the equalizer tap length from 3 to 5 showed reasonable performance improvement.
2. Diminishing returns from improving the equalizer tap length above 5.
3. The results are poorer compared to the AWGN no multipath results. This is due to the noise amplification (see the frequency response above) by the zero forcing equalization filter.
Next step is to discuss the zero forcing equalizer in the presence of transmit pulse shaping and then move on to minimum mean square error equalizer.
Related posts
- BER for BPSK in ISI channel with MMSE equalization
- MIMO with Zero Forcing Successive Interference Cancellation equalizer
- MIMO with Zero Forcing equalizer
- BER with Matched Filtering
- MIMO with MMSE SIC and optimal ordering
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Wow~~ New topic, I like it~!
thank you so much!!
Fantastic work! really. Can’t wait for MMSE equalization (I think that would be next)
Krishna, would you please help me with the two questions I have with respect to your posts on OFDM and EGC. I have provided the link below
For OFDM:
http://www.dsplog.com/2008/08/26/ofdm-rayleigh-channel-ber-bpsk/#comment-15910
For EGC:
http://www.dsplog.com/2008/09/19/equal-gain-combining/#comment-16350
I’d appreciate your help
@communications_engineer: Thanks. Hope my reply to both the comments addressed some of your concerns. Good luck.
Very nice work , AMAZING
Keep Goning
@Ayhem: Thanks
Hi Krishna
Thank you for very nice post
May I ask you about the basic question
From the script below
10^(-Eb_N0_dB(ii)/20)*n; % additive white gaussian noise
I know that
10log10(A_power) = A_power dB —- (1)
20log10(A_voltage) = A_voltage dB —– (2)
when we try to convert Eb_N0 into dB we use formula (1)
10log10(Eb_N0) = Eb_N0_dB
Why does the division factor equal to 20 ?
Thanks
@Puripong: We want to convert scale the noise voltage by a factor corresponding to Eb_N0_dB. To convert dB to voltage we have the 1/20 term (as you have pointed out in (2)). Helps?
You mean….
We want to convert Eb_N0_dB into noise voltage.
sqrt(voltage) = Eb_N0_dB
voltage = Eb_N0_dB^2
10log10(Eb_N0_dB^2) = 20log10(Eb_N0_dB)
So, we have the division factor equal to 20
Ohhh, Surely I missed.
Thank you very much
@Puripong: Glad to help
Thanks A Lot For This
Thanks for your post!
When I saw some old posts, many red “X” appear where the formula or symbol should be. How can I solve it ? Is something wrong with my browser?
hi,
In your code,
ht = [0.4 0.9 0.3];
I tried another value like ht = [0.6 0.9 0.3];
then BER increased dramatically, it doesn;t make sense
Can you tell me what’s the problem? THX
Thanks for your post!
When I saw some old posts, many red “X” appear where the formula or symbol should be. How can I solve it ? Is something wrong with my browser?
@rajesh: Can you please list the posts where you saw this error. This is not expected. Anyhow once you give the list, I will take a look.
It’s a useful one for me
please send me multi-user detectors in ds-cdma
hello, Krishna Sankar, I have a question on ZF for ISI channel.
As your example, I have ISI channel with three taps [h1 h2 h3], then the signal [S1 S2 S3 S4 S5 S6 S7 ...] passes through the channel. At the receiver, the received signal denoted by [R1 R2 R3 R4 R5 R6 R7...]. For example, I want to detect symbol R6, we can readily find that R6=h1*S6+h2*S5+h3*S4, then this signal passes through ZF equalizer with coefficients [C1 C2 C3]. Then the estimated symbol for R6 will be :
^R6=C1*R6+C2*R5+C3*R4=C1h1S6+(C1h2+C1h1)S5+(C1h3+C2h2+C3h1)S4+(C2h3+C3h2)S3+c3h3S2
Why it’s different from your derivation results. Can you tell where I’m wrong.
Hi Krishna,
Request you to please explain the below mentioned part of your code.
K=4; What is the significance of this K and why is it equal to 4?
for kk = 1:K
L = length(ht);
hM = toeplitz([ht([2:end]) zeros(1,2*kk+1-L+1)], [ ht([2:-1:1]) zeros(1,2*kk+1-L+1) ]);
d = zeros(1,2*kk+1);
d(kk+1) = 1;
c = [inv(hM)*d.'].’;
% mathched filter
yFilt = conv(y,c);
yFilt = yFilt(kk+2:end);
yFilt = conv(yFilt,ones(1,1)); % convolution
ySamp = yFilt(1:1:N); % sampling at time T
Can we not perform the above operations using just conv or filter functions in Matlab?
Please provide your comments.
Thanks
Ishwinder
Hi Krishna,
One question about the channel coeffecients. They are real number as I we find here. But in case of Rayleigh channel they are complex where both real are imaginary parts are samples from Gaussain distribution. My question is if we consider a static ISI channel with complex coeffcients, is there anything wrong with it?Since the co-efficients refer to the gain of each tap, I feel complex values are also possible where the imaginary parts refers to the phase.
Waiting for your response.
Ahmed