In electromathematics two important functions play a major role in determining the system response to an arbitrary input. Indeed it's only one function which is called as the Transfer function of the system. From the transfer function of the system we can find out the characteristic equation. It's necessary that you should understand these functions intuitively before proceeding to solve problems based on this. I don't want to complicate things so I want to keep it simple with some basic examples which I will discuss later on. But before that you should know that whenever you subject a system to an input it has two responses or two outputs.
One is the natural response and other one is the forced response. In the beginning of your education such as in 12th grade or in some high school what you had learnt is the forced response alone (i.e. response of the system after a great deal of time, in other words t--> infinty. ) But in reality the system don't respond to the input immediately. It takes some time to reach the final state.
Let's take a capacitor and a resistor in series. When you apply an input with zero frequency (i.e. dc) then capacitor blocks dc and no current flows and hence the entire drop appears across the capacitor. But this don't take place instantaneously because if you take the capacitor equation.
i (t) = C dv/dt
This means that when the voltage at the input suddenly raises from 0 V to some 5V (say) at t=0 then the differentiation at that time t=0 becomes infinite becomes of the infinite slope. But infinite current is impossible as it requires infinite power/ energy. So current doesn't change instantaneously. If you solve the above differential equation (refer William Hayt "Engineering Circuit Analysis" for detailed explanation of natural and forced response) you will get
i(t) = A e^(-t/RC).
Where 'A' is some constant. After a certain time i(t) dies out and when t=infinite the current becomes zero.
So we can say that the response consists of two terms -
1) natural response
2) forced response
Output= forced response + natural response.
Take another situation.
There are certain system where the natural response doesn't die out at all. In other words the power in the exponential term is positive or zero. One such system is a special case of underdamped system (refer William Hayt "Engineering Circuit Analysis" for detailed explanation ).
There is another way of seeing the natural response system. It's through the transfer function of the system. We have to note that the natural response of the system is dependent only on the circuit elements and not on the input or the output. As I have told in one of my previous posts Characteristic equation (or the laplace transform ) of the system is any easy way of seeing the natural response of the system. Instead of some exponential term we deal with the algebraic equations. All we care about is the power of the exponential. (see my first post if you don't understand what I have said)
If the system characteristic equation is s + a = 0 i.e. we have pole at s = -a then the natural response of the system is an exponential with power having the term "-a". So that if a is positive then the system is stable and if it is negative or zero then the system becomes unstable. In other words if the pole of the characteristic equation is positive or equal to zero then the system becomes unstable. This is the reason why we say that we should not have a pole on the right hand side of the jw axis.
There is also another use of the transfer function. It's used to find out the system response for any input. It can be shown that the system function is equal to the transfer function of the system(The proof of this can be found in the same book that I had mentioned before). So for example if the Transfer function is
T(s) = 1 / (s +a + jw)(s+ a -jw). The characteristic equation is the denominator and the numerator doesn't matter at all if you consider the stability due to the natural response of the system.
Vout(s) / Vin (s) = 1/( s + a + jw)(s + a -jw)
It's necessary that whenever you have a complex pole you should have a conjugate of the that pole. This is because we can't have imaginary terms in the time domain of the natural response and it will be real only when we have conjugate. This is because of the euler's formula for cos x and sin x where it contains conjugate terms. Again I can't say much about this ( refer the same book).
Okay coming back to the output.
Vout(s) = [ 1/(s + a + jw)(s + a -jw)] Vin(s).
It can be shown that for a low pass filter this term 'w' should be near 0 and similarly for high pass it should be near infinite. Hence for greater selectivity ( in other words jw term should dominate more than the term 'a') we should have 'a' to be very small. In other words poles should be near the jw axis. Or else the term jw has no effect at all and we can't say whether it is low pass or high pass.
Note that when we have a=0 the output becomes infinite when s = jw(i.e. poles lie on the jw axis). It's also due to the fact that the exponential term has non-negative power (since it's power is -a). Hence the system becomes unstable.
Also if we consider the real part is equal to wi/Q for a second order transfer function where wi is the distance of the pole from the origin then it has high selectivity if real part 'a' is small i.e. Q is large. This is the reason for having this term Q in the denominator of wi for the expression of a. You can also verify this graphically. When Q is high the curve is steeper.
If you want to learn in depth about Laplace transform refer "Engineering Circuit Analysis" by William Hayt. That's the best book for intuitive understanding of Electromathematics.
Till then bye
iVenky.
One is the natural response and other one is the forced response. In the beginning of your education such as in 12th grade or in some high school what you had learnt is the forced response alone (i.e. response of the system after a great deal of time, in other words t--> infinty. ) But in reality the system don't respond to the input immediately. It takes some time to reach the final state.
Let's take a capacitor and a resistor in series. When you apply an input with zero frequency (i.e. dc) then capacitor blocks dc and no current flows and hence the entire drop appears across the capacitor. But this don't take place instantaneously because if you take the capacitor equation.
i (t) = C dv/dt
This means that when the voltage at the input suddenly raises from 0 V to some 5V (say) at t=0 then the differentiation at that time t=0 becomes infinite becomes of the infinite slope. But infinite current is impossible as it requires infinite power/ energy. So current doesn't change instantaneously. If you solve the above differential equation (refer William Hayt "Engineering Circuit Analysis" for detailed explanation of natural and forced response) you will get
i(t) = A e^(-t/RC).
Where 'A' is some constant. After a certain time i(t) dies out and when t=infinite the current becomes zero.
So we can say that the response consists of two terms -
1) natural response
2) forced response
Output= forced response + natural response.
Take another situation.
There are certain system where the natural response doesn't die out at all. In other words the power in the exponential term is positive or zero. One such system is a special case of underdamped system (refer William Hayt "Engineering Circuit Analysis" for detailed explanation ).
There is another way of seeing the natural response system. It's through the transfer function of the system. We have to note that the natural response of the system is dependent only on the circuit elements and not on the input or the output. As I have told in one of my previous posts Characteristic equation (or the laplace transform ) of the system is any easy way of seeing the natural response of the system. Instead of some exponential term we deal with the algebraic equations. All we care about is the power of the exponential. (see my first post if you don't understand what I have said)
If the system characteristic equation is s + a = 0 i.e. we have pole at s = -a then the natural response of the system is an exponential with power having the term "-a". So that if a is positive then the system is stable and if it is negative or zero then the system becomes unstable. In other words if the pole of the characteristic equation is positive or equal to zero then the system becomes unstable. This is the reason why we say that we should not have a pole on the right hand side of the jw axis.
There is also another use of the transfer function. It's used to find out the system response for any input. It can be shown that the system function is equal to the transfer function of the system(The proof of this can be found in the same book that I had mentioned before). So for example if the Transfer function is
T(s) = 1 / (s +a + jw)(s+ a -jw). The characteristic equation is the denominator and the numerator doesn't matter at all if you consider the stability due to the natural response of the system.
Vout(s) / Vin (s) = 1/( s + a + jw)(s + a -jw)
It's necessary that whenever you have a complex pole you should have a conjugate of the that pole. This is because we can't have imaginary terms in the time domain of the natural response and it will be real only when we have conjugate. This is because of the euler's formula for cos x and sin x where it contains conjugate terms. Again I can't say much about this ( refer the same book).
Okay coming back to the output.
Vout(s) = [ 1/(s + a + jw)(s + a -jw)] Vin(s).
It can be shown that for a low pass filter this term 'w' should be near 0 and similarly for high pass it should be near infinite. Hence for greater selectivity ( in other words jw term should dominate more than the term 'a') we should have 'a' to be very small. In other words poles should be near the jw axis. Or else the term jw has no effect at all and we can't say whether it is low pass or high pass.
Note that when we have a=0 the output becomes infinite when s = jw(i.e. poles lie on the jw axis). It's also due to the fact that the exponential term has non-negative power (since it's power is -a). Hence the system becomes unstable.
Also if we consider the real part is equal to wi/Q for a second order transfer function where wi is the distance of the pole from the origin then it has high selectivity if real part 'a' is small i.e. Q is large. This is the reason for having this term Q in the denominator of wi for the expression of a. You can also verify this graphically. When Q is high the curve is steeper.
If you want to learn in depth about Laplace transform refer "Engineering Circuit Analysis" by William Hayt. That's the best book for intuitive understanding of Electromathematics.
Till then bye
iVenky.
