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Suppose you are given a system with transfer function

$$H(z)=\frac{(1-3z^{-1})(1-7z^{-1})}{(1-4z^{-1})(1-6z^{-1})}$$$$H(z)=\frac{(1-3z^{-1})(1-7z^{-1})}{(1-4z^{-1})(1-6z^{-1})}$$

### Poles

Poles are the values of z$$z$$ for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?

Look here, what values of z$$z$$ will turn the transfer function tend to infinity? Obviously itsit's $$z= 4$$and 6 and $$z=6$$, because if you put z equalslet $$z$$ equal 4 or 6, the denominator will be zero, which means the transfer function will tend to infinity.

### Zeros

Again, zerosZeros are the values of z for which the transfer function will be zero. As you have guessed correctly, zeros come from numerator. In this case, zeros are $$z= 3$$ and $$z=7$$, cause if you put $$z= 3$$ or $$z=7$$, the numerator will be zero, that means the whole transfer function will be zero.

So here poles are $$z=4$$ and $$z=6$$, and zeros are $$z=3$$ and $$z=7$$.

### Pole-Zero Plot

About finding the Pole zero plot, you draw a complex plane. Then you put the values of poles as 'X' marks and zeros as 'O' marks. Here I took the liberty of drawing the pole zero plot of the system . Pole zero plot of the above system:

So, for low pass filter, you find out the transfer function, then the poles and zeros.

Suppose you are given a system with transfer function

$$H(z)=\frac{(1-3z^{-1})(1-7z^{-1})}{(1-4z^{-1})(1-6z^{-1})}$$

### Poles

Poles are the values of z for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?

Look here, what values of z will turn the transfer function tend to infinity? Obviously its $$z= 4$$and 6, because if you put z equals 4 or 6, the denominator will be zero, which means the transfer function will tend to infinity.

### Zeros

Again, zeros are the values of z for which the transfer function will be zero. As you have guessed correctly, zeros come from numerator. In this case, zeros are $$z= 3$$ and $$z=7$$, cause if you put $$z= 3$$ or $$z=7$$, the numerator will be zero, that means the whole transfer function will be zero.

So here poles are $$z=4$$ and $$z=6$$, and zeros are $$z=3$$ and $$z=7$$.

### Pole-Zero Plot

About finding the Pole zero plot, you draw a complex plane. Then you put the values of poles as 'X' marks and zeros as 'O' marks. Here I took the liberty of drawing the pole zero plot of the system . Pole zero plot of the above system

So, for low pass filter, you find out the transfer function, then the poles and zeros.

Suppose you are given a system with transfer function

$$H(z)=\frac{(1-3z^{-1})(1-7z^{-1})}{(1-4z^{-1})(1-6z^{-1})}$$

### Poles

Poles are the values of $$z$$ for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?

Look here, what values of $$z$$ will turn the transfer function tend to infinity? Obviously it's $$z= 4$$ and $$z=6$$, because if you let $$z$$ equal 4 or 6, the denominator will be zero, which means the transfer function will tend to infinity.

### Zeros

Zeros are the values of z for which the transfer function will be zero. As you have guessed correctly, zeros come from numerator. In this case, zeros are $$z= 3$$ and $$z=7$$, cause if you put $$z= 3$$ or $$z=7$$, the numerator will be zero, that means the whole transfer function will be zero.

So here poles are $$z=4$$ and $$z=6$$, and zeros are $$z=3$$ and $$z=7$$.

### Pole-Zero Plot

About finding the Pole zero plot, you draw a complex plane. Then you put the values of poles as 'X' marks and zeros as 'O' marks. Here I took the liberty of drawing the pole zero plot of the system:

So, for low pass filter, you find out the transfer function, then the poles and zeros.

3 latex markup edits for readability

Suppose you are given a system with transfer function

H(z)=(1-3z-1)(1-7z-1) / (1-4z-1)(1-6z-1) $$H(z)=\frac{(1-3z^{-1})(1-7z^{-1})}{(1-4z^{-1})(1-6z^{-1})}$$

### Poles

Poles are the values of z for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?

Look here, what values of z will turn the transfer function tend to infinity? Obviously its z= 4 and$$z= 4$$and 6, because if you put z equals 4 or 6, the denominator will be zero, which means the transfer function will tend to infinity.

### Zeros

Again, zeros are the values of z for which the transfer function will be zero. As you have guessed correctly, zeros come from numerator. In this case, zeros are z= 3$$z= 3$$ and 7$$z=7$$, cause if you put z= 3$$z= 3$$ or 7$$z=7$$, the numerator will be zero, that means the whole transfer function will be zero.

So here poles are z=4$$z=4$$ and 6$$z=6$$, and zeros are z=3$$z=3$$ and 7$$z=7$$.

### Pole-Zero Plot

About finding the Pole zero plot, you draw a complex plane. Then you put the values of poles as 'X' marks and zeros as 'O' marks. Here I took the liberty of drawing the pole zero plot of the system . Pole zero plot of the above system

So, for low pass filter, you find out the transfer function, then the poles and zeros.

Suppose you are given a system with transfer function

H(z)=(1-3z-1)(1-7z-1) / (1-4z-1)(1-6z-1)

### Poles

Poles are the values of z for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?

Look here, what values of z will turn the transfer function tend to infinity? Obviously its z= 4 and 6, because if you put z equals 4 or 6, the denominator will be zero, which means the transfer function will tend to infinity.

### Zeros

Again, zeros are the values of z for which the transfer function will be zero. As you have guessed correctly, zeros come from numerator. In this case, zeros are z= 3 and 7, cause if you put z= 3 or 7, the numerator will be zero, that means the whole transfer function will be zero.

So here poles are z=4 and 6, and zeros are z=3 and 7.

### Pole-Zero Plot

About finding the Pole zero plot, you draw a complex plane. Then you put the values of poles as 'X' marks and zeros as 'O' marks. Here I took the liberty of drawing the pole zero plot of the system . Pole zero plot of the above system

So, for low pass filter, you find out the transfer function, then the poles and zeros.

Suppose you are given a system with transfer function

$$H(z)=\frac{(1-3z^{-1})(1-7z^{-1})}{(1-4z^{-1})(1-6z^{-1})}$$

### Poles

Poles are the values of z for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?

Look here, what values of z will turn the transfer function tend to infinity? Obviously its $$z= 4$$and 6, because if you put z equals 4 or 6, the denominator will be zero, which means the transfer function will tend to infinity.

### Zeros

Again, zeros are the values of z for which the transfer function will be zero. As you have guessed correctly, zeros come from numerator. In this case, zeros are $$z= 3$$ and $$z=7$$, cause if you put $$z= 3$$ or $$z=7$$, the numerator will be zero, that means the whole transfer function will be zero.

So here poles are $$z=4$$ and $$z=6$$, and zeros are $$z=3$$ and $$z=7$$.

### Pole-Zero Plot

About finding the Pole zero plot, you draw a complex plane. Then you put the values of poles as 'X' marks and zeros as 'O' marks. Here I took the liberty of drawing the pole zero plot of the system . Pole zero plot of the above system

So, for low pass filter, you find out the transfer function, then the poles and zeros.

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Suppose you are given a system with transfer function

H(z)=(1-3z-1)(1-7z-1) / (1-4z-1)(1-6z-1)

### Poles

Poles are the values of z for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?

Look here, what values of z will turn the transfer function tend to infinity? Obviously its z= 4 and 6, because if you put z equals 4 or 6, the denominator will be zero, which means the transfer function will tend to infinity.

### Zeros

Again, zeros are the values of z for which the transfer function will be zero. As you have guessed correctly, zeros come from numerator. In this case, zeros are z= 3 and 7, cause if you put z= 3 or 7, the numerator will be zero, that means the whole transfer function will be zero.

So here poles are z=4 and 6, and zeros are z=3 and 7.

### Pole-Zero Plot

About finding the Pole zero plot, you draw a complex plane. Then you put the values of poles as 'X' marks and zeros as 'O' marks. Here I took the liberty of drawing the pole zero plot of the system . Pole zero plot of the above system

So, for low pass filter, you find out the transfer function, then the poles and zeros.

Suppose you are given a system with transfer function

H(z)=(1-3z-1)(1-7z-1) / (1-4z-1)(1-6z-1)

### Poles

Poles are the values of z for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?

Look here, what values of z will turn the transfer function tend to infinity? Obviously its z= 4 and 6, because if you put z equals 4 or 6, the denominator will be zero, which means the transfer function will tend to infinity.

### Zeros

Again, zeros are the values of z for which the transfer function will be zero. As you have guessed correctly, zeros come from numerator. In this case, zeros are z= 3 and 7, cause if you put z= 3 or 7, the numerator will be zero, that means the whole transfer function will be zero.

So here poles are z=4 and 6, and zeros are z=3 and 7.

### Pole-Zero Plot

About finding the Pole zero plot, you draw a complex plane. Then you put the values of poles as 'X' marks and zeros as 'O' marks.

So, for low pass filter, you find out the transfer function, then the poles and zeros.

Suppose you are given a system with transfer function

H(z)=(1-3z-1)(1-7z-1) / (1-4z-1)(1-6z-1)

### Poles

Poles are the values of z for which the entire function will be infinity or undefined. So, they will be the roots of the denominators, right?

Look here, what values of z will turn the transfer function tend to infinity? Obviously its z= 4 and 6, because if you put z equals 4 or 6, the denominator will be zero, which means the transfer function will tend to infinity.

### Zeros

Again, zeros are the values of z for which the transfer function will be zero. As you have guessed correctly, zeros come from numerator. In this case, zeros are z= 3 and 7, cause if you put z= 3 or 7, the numerator will be zero, that means the whole transfer function will be zero.

So here poles are z=4 and 6, and zeros are z=3 and 7.

### Pole-Zero Plot

About finding the Pole zero plot, you draw a complex plane. Then you put the values of poles as 'X' marks and zeros as 'O' marks. Here I took the liberty of drawing the pole zero plot of the system . Pole zero plot of the above system

So, for low pass filter, you find out the transfer function, then the poles and zeros.

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