pole zero simulation result

[sj95]

Hi all:

spice simulation tell me

poles ( hertz)
real imag
-1.59158 0.
-79.5759k 0.

zeros ( hertz)
real imag
0. 355.881
0. -355.881


that says there is 2 zero at 355.881 Hz? how about the minus sign ? what's its meaning?

thanks

- - - Updated - - -

And


poles ( hertz)
real imag
-837.259k 0.
-2.01350x 2.29293x
-2.01350x -2.29293x


that say pole 1 at 837.259kHz ?
But how about P2 and P3 , that are at 2.01350xHZ ?

tks

 

[LvW]

First example:
Both zeros are symmetrically located on the imaginary axis of the s-plane. This is a typical property of a bandstop function.

Second example:
It is a third-order system with one real pole and two conjugate-complex poles. The pole frequency of this pole pair is the MAGNITUDE - composed of the real and imaginary part ( SQRT(R^2 + Im^2) ) .

 

[gasingh]

heya Sj95

Let me elaborate a bit on what LvW already said ,

To understand the physical meaning let us first agree that we are giving a sinusoidal input to the system say Asinwt and measuring the amplitude and phase shift the output say B(w).sin(wt + Phi(w) )

->Clearly the gain is B(w)/A and phase shift produced by the system is pHi(w)

->Note that both gain and phase shift can be functions of the frequency w , as they usually are and fully described by the transfer function

H(w) = K.{ ( jw - z1 ) ( jw - z2) ...} / {(jw -p1 )(jw-p2)... }


-> Now our objective is to move along jw axis and measure the gain and the phase shift provided by the system

"While there is nothing real like negative frequency , or even negative numbers or even complex numbers , but by extending our mathematics to these domains we can correctly compute the behavior of the system to the real frequencies " , so when you see that there is a zero at -jw1 and -jw2, although you never can give a input with frequency -w1 , the mathematics still allows you to compute the output for any frequency w > 0 , so the zero at negative frequency has a meaning in a way that it effects the amplitude and phase shift that your system provides at any real frequency "



What transfer function above gives you is the system response (Amp and Phase ) to input A.e^jwt ( a complex number ) , but if you know the response at A.e^jwt and A.e^-jwt , you can calculate the response by superimposition for A.sinwt ( your real frequency)

as sinwt = (e^jwt - e^-jwt)/2

It is now easy to see why your first system is a bandstop , For complex Input A.e^j(0. 355.881 )t the output(complex) is zero and for the complex Input A.e^j(-0. 355.881 )t also the output (complex) is zero , if you now combine the two inputs and do the superposition the output for Asinwt (real ) is also zero

I know its not very clear but I have tried my best .

 

[LvW]

Hi Gasingh,

nice sentence:

"While there is nothing real like negative frequency , or even negative numbers or even complex numbers , but by extending our mathematics to these domains we can correctly compute the behavior of the system to the real frequencies " , so when you see that there is a zero at -jw1 and -jw2, although you never can give a input with frequency -w1 , the mathematics still allows you to compute the output for any frequency w > 0 , so the zero at negative frequency has a meaning in a way that it effects the amplitude and phase shift that your system provides at any real frequency "

As you have used quotation marks I assume somebody else has created this sentence. What/who is the source?

 

[gasingh]

Hi Gasingh,

nice sentence:

"While there is nothing real like negative frequency , or even negative numbers or even complex numbers , but by extending our mathematics to these domains we can correctly compute the behavior of the system to the real frequencies " , so when you see that there is a zero at -jw1 and -jw2, although you never can give a input with frequency -w1 , the mathematics still allows you to compute the output for any frequency w > 0 , so the zero at negative frequency has a meaning in a way that it effects the amplitude and phase shift that your system provides at any real frequency "

As you have used quotation marks I assume somebody else has created this sentence. What/who is the source?

Thanks for Liking , With all humbleness, its my own line .