DSP in SciLAB

[ΔЅịMø]

You have also to simplify this by converting a0 to be equal to 1 to make this into direct form 2.

           3 – 5z^-1 + 2z^-3
H(z) = ---------------------------
       6 – 29z^-1 – 7z^-2 + 10z^-3

           3 – 5z^-1 + 2z^-3         1/6
H(z) = --------------------------- x ---
       6 – 29z^-1 – 7z^-2 + 10z^-3   1/6

           0.5 – 0.833z^-1 + 0.333z^-3
H(z) = -------------------------------------
       1 – 4.833z^-1 – 1.167z^-2 + 1.667z^-3

a1 = 29/6
a2 = 7/6
a3 = 10/6
b0 = 0.5
b1= 5/6
b2 = 0
b3 = 2/6

Medjo nakuha ko po ito sir. Kailangan po talaga na 1 yung a0 to be considered as direct form II.



Dun lang po ako medjo naguluhan sa Reply No. 38. Pasensya na po sir.

[ΔЅịMø]

Next step na pos ir is this one:

                      W(z)
X(z) --> H1(z) --- > H2(z) ---> Y(z)

 
               W(z)
H1(z)=---------- =  0.5 – 0.833z-1 + 0.333z-3
           X(z)

Applying cross multiplication:

w[n]=0.5x[n]-0.833x[n-1]+0.333x[n-2]

and

                Y(z)                 1
H2(z)=---------- = --------------------------------------
                W(z)        1 – 4.833z-1 – 1.167z-2 + 1.667z-3

 
Applying cross multiplication:

w[n]=1y[n]-4.833y[n-1]-1.167y[n-2]+1.167y[n-3]

Tama po ba sir?

[zer0w1ng]

No need to go through the process of conversion.
I just showed it to prove that direct form 2 is derived from direct form1.

You can just simply replace the values of a1, a2, a3, b0, b1, b2 and b3 on the direct form 2 diagram.

a1 = 29/6  = 4.833
a2 = 7/6   = 1.167
a3 = 10/6  = 1.667
b0 = 3/5   = 0.5
b1= 5/6    = 0.833
b2 = 0
b3 = 2/6   = 0.333

  o----->(+)----------o--| b0 >-->(+)----p
          ^           |            ^   
          |          1/z           |   
          |           |            |   
          o--< -a1 |--o--| b1 >----o   
          ^           |            ^   
          |          1/z           |   
          |           |            |   
          o--< -a2 |--o--| b2 >----o   
          ^           |            ^   
          |          1/z           |   
          |           |            |   
          o--< -a3 |--o--| b3 >----o   

[ΔЅịMø]

Here was the diagram in SciLAB's XCOS po sir:








Kaya lang po sir, NO OUTPUT po sya.

[zer0w1ng]

It may have no output because it may have been filtered out.
As you know, this is an IIR filter and either it increases or decreases the response depending on the frequency of the input.

As stated by others too, an impulse input maybe better to check your filter for oscillations or response.

Try changing the frequency of the input and see the result.

[ΔЅịMø]

Pinalitan ko na po yung magnitude from 1 to 20, 50, 100, 500 sir. But, wala pong output.

But, if ito pong diagram na to yung gagamitin ko sir






meron po syang output. About 2 po yung magnitude nya sir.

[zer0w1ng]

Why are you changing the magnitude?
It should be the *FREQUENCY*.

Your above diagram is wrong that is why there is a difference in output.

Please note that having *NO OUPUT* might be the *CORRECT* response of your filter because it may be removing that specific frequency on your input.

[ΔЅịMø]

Sorry sir. Frequency pala. Mali lang po.

Possible din po pala na wala pong output na ipapalabas ang filter sir?

[ΔЅịMø]

As i simulate the diagram in cascaded output sir, here is the result with a frequency=80.






Legend:

GREEN COLORED WAVE = OUTPUT
BLACK COLORED WAVE = INPUT

[ΔЅịMø]

Sir zero, tapos ko na po. Salamat po sa tulong and patience. +1 po.

[your most RIDICULOUS man :)]

Genius!

[ΔЅịMø]

Genius!

At binuhay ulit.

Di ko lang i-la-lock para kung amy magtatanong pa about SciLAB, pwede sila dito.

[your most RIDICULOUS man :)]

galeng talaga..

[centOS]

very nice sir galing mo lalo na si sir Zero