Why am I getting this error "??? error using ==> times matrix dimensions must agree" in matlab?

Debajyoti (view profile)

on 21 Apr 2013
Accepted Answer by Cedric Wannaz

Cedric Wannaz (view profile)

design a FIR low pass filter using Kaiser window

```a1phap=0.1
```
```a1phas=44
```
```ws=30
```
```wp=20
```
```wsf=100
```
```B=ws-wp
```
```wc=0.5*(ws+wp)
```
```wcr=wc*2*pi/wsf
```
```D=(a1phas-7.95)/14.36
```
```N=ceil((wsf*D/B)+1)
```
```a1pha=(N-1)/2
```
```gamma=(.5842*(a1phas-21).^(0.4)+0.07886*(a1phas-21))
```
```n=0:1:N-1
```
```hd=sin(wcr*(n-a1pha))./(pi*(n-a1pha))
```
```hd(a1pha+1)=0.5
```
```wk=(kaiser(N,gamma))
```
```hn=hd.*wk     % *(HERE I AM GETTING THIS ERROR)*
```
```W=85:5:120
```
```h=freqz(hn,1,W)
```
```plot(W/pi,20*log10(abs(h)))
```

those two matrices have the same dimensions of 27..still why this error?

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Cedric Wannaz (view profile)

Answer by Cedric Wannaz

Cedric Wannaz (view profile)

on 21 Apr 2013

You have to transpose either wk or hd, because one is a column vector and the other a row vector. E.g.

` hn = hd .* wk.'`

the cyclist (view profile)

Answer by the cyclist

the cyclist (view profile)

on 21 Apr 2013
Edited by the cyclist

the cyclist (view profile)

on 21 Apr 2013

I don't have the Signal Processing Toolbox, so I can't execute the kaiser() function. However, the documentation indicates that it will output a column vector. Your vector hd is a row vector, so it does not actually have the same dimension. You need to transpose one of them.

Chandra Shekhar (view profile)

Answer by Chandra Shekhar

Chandra Shekhar (view profile)

on 21 Apr 2013
Edited by Chandra Shekhar

Chandra Shekhar (view profile)

on 21 Apr 2013

Don't use dot product,edit that line like this

hn=hd*wk;

but you will get single value after multiplying,but 'freqz' function requires vector as a input.

if you need vector then do like this, you will get waveform.

hn=hd. * wk ' ;

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