why Pole cause -6db/oct in Bode plot?

[020170]

6db/oct

some book tell me "pole is -6db/oct and Zero is 6db/oct"

but why?

I know that the decreasing reactance of the capacitor with increasing Frequency.

but How did you know exactly that Pole makes -6db/octave roll-off by decreasing reactance?

I understood that capacitor means open circuit in "pole", and inductor means short circuit in "ZERO".

but I don't understand why these properties make -6 or +6 db/oct roll-off in bode plot.

thanks.

 

[ronyutp]

pole 6db

Lets see:

I think you know that that transfer function of system with a pole could be written:

G(s)=K/(s+p)

or in frequency

G(jω) =K/√(ω^2 +p^2)

where p is the pole and K is a constant value. -6db/oct means that when frequency is too big ( ω-> ∞ ) your circuit has an attenuation of 6db/oct.

 

[VVV]

bode plot,pole shift

First of all, -6dB/oct is an approximation. The correct number is -20dB/decade.

How do we get -20dB/dec?
Let's consider the transfer function: H(s)=K/(1+s/ωp).
For frequency response, you really replace s by jω.

H(jω)=K/(1+jω/ωp)

As you can see, as long as ω is very small (ω<<ωp, low frequency), the transfer function is approximately equal to the constant K, since the denominator is approximately 1.

When you reach the pole frequency, the gain in the Bode plot is 20*log|H(jω)|=20*log(K/√2)=20*logK+20*log(1/√2)
The attenuation then is seen to be 20*log(1/√2)= -3.01dB. This is the real-world attenuation at exactly the pole frequency, which is why we determine the pole frequency by finding when the attenuation is -3dB. In an ideal Bode plot, the pole frequency is just a corner, in real world it is rounded, the difference being -3dB. So by finding the -3dB point, we really find the ideal corner frequency.

Now, well above the pole frequency (ω>>ωp, high frequency), the denominator is approximately jω/ωp. Thus, the gain is 20*log(K/|jω/ωp|). As you can see, a frequency increase of 10 (a decade) means 20*log(K/|j*10*ω/ωp|)=20*logK+20*log(1/|jω/ωp|)+20*log(1/10). But 20*log(1/10)=-20dB, thus you get -20dB/dec.

Or, if you prefer per octave: 20*log(K/|j*2*ω/ωp|)=20*logK+20*log(1/|jω/ωp|)+20*log(1/2). But 20*log(1/2)=-6.02dB. This is your -6dB/oct.

Now consider the RC integrator in the figure. Its transfer function is H(s)=Uo(s)/Ui(s). Or, for Bode plots, H(jω)=Uo(jω)/Ui(jω).
The circuit can be analyzed as a simple voltage divider, using R1 and the impedance of the cap. The impedance of the cap is Xc=1/jωC.

Then, Uo(jω)=Ui(jω)*(1/jωC)/((1/jωC)+R)

With that, H(jω)=Uo(jω)/Hi(jω)=(1/jωC)/((1/jωC)+R)=1/(1+jωCR)

Noting that the time constant is τ=RC, the pole frequency is fp=1/(2πRC), or ωp=1/RC.
So the transfer function becomes: H(jω)=Uo(jω)/Ui(jω)=1/(1+jω/ωp).
In other words, the constant K is 1, which means that at low frequency the output voltage is approximately equal to the input voltage, that is, the cap plays no role. As the frequency increases, you reach the pole frequency. The rest is as explained above.

The reason why the cap shapes the Bode plot the way it does is because its impedance decreases linearly with increasing frequency.

 

[020170]

bode plot 6db/octave 20db/decade

thanks!

finally, I understand completly what you said!

thanks!

I appreciate your help!

 

[Kral]

+octave+bode

A postript to VVVs reply: The exact value for "6db" is 20[log(2)], which is approximately equal to 6.020599913.

 

[sandeep_torgal]

how much db/oct rolloff does 1 pole represent

First of all, -6dB/oct is an approximation. The correct number is -20dB/decade.

How do we get -20dB/dec?
Let's consider the transfer function: H(s)=K/(1+s/ωp).
For frequency response, you really replace s by jω.

H(jω)=K/(1+jω/ωp)

As you can see, as long as ω is very small (ω<<ωp, low frequency), the transfer function is approximately equal to the constant K, since the denominator is approximately 1.

When you reach the pole frequency, the gain in the Bode plot is 20*log|H(jω)|=20*log(K/√2)=20*logK+20*log(1/√2)

I didnt get , How did 20log(1/√2) come....

Regards,
Sandeep

 

[FvM]

6 db octave bode plot

1/√2 = |1/(1 + j)| = 1/√(1² + 1²) for ω = ωp

If you prefer a geometrical explanation: For ω = ωp Uc1 and Ur1 in the above low-pass have same magnitude but 90° phase shift, thus the vectorial voltage sum has a √2 larger magnitude. You get 1/√2 when you calculate the attenuation of the R/C network.

 

[sandeep_torgal]

poles on bode plot

Hi VVV,
I overlooked |H(jω)| and didnt realise that it was the magnitude.

Regards,
Sandeep

 

[shou]

RC filter
Av=-20*log(SQRT(1+(ω/ω0)^2)=-20*log(ω/ω0)
ω/ω0=10 ⇒　Av＝-20*log10=-20dB
ω/ω0=2 ⇒　Av＝-20*log2 ＝-6dB