1st order filters-non inverters

8/6/2019 1st Order Filters-non Inverters

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Federal University of Santa Maria

Technology Center

Electrical Engineering Department

UFSM

1st Order Non-Inverter Filters Analysis

Low-pass and High-pass

Signal Processing

Author:

Tiago Oliveira WeberProfessor:

Alexandre Campos

Santa Maria, RS, BrazilSeptember 29th



Introduction

On the present paper, a analysis of non inverters first order filters will be presented.

Both low-pass and high-pass filters will be studied, considering their circuit, block diagramand bode diagram.

General First Order Filters

For a 1st order LTI system:

H (s) =a1.s + a0

b1.s + b0

To project a low-pass filter a1 must be equal to zero.

H (s) =a0

b1.s + b0

H (s) =a0

b0.

b0

b1

s + b0

b1

It’s useful to see this transfer function as:

H (s) = H 0.w0

s + w0

To project a high-pass filter a0 must be equal to zero

H (s) =a0

b1.s + b0

and this equation is better visualized as

H (s) = H inf .s

s + w0

Low-pass Non-Inverter Filter

The first part has a transfer function equal to:

G(s)1 =1

s.R.C

1 + 1s.R.C

G(s)1 =1

1 + s.R.C

The second part has a transfer function as following:

Signal Processing 1 Tiago Oliveira Weber



G( jw) =H 0

1 + ( w

w0)2− j

H 0. w

w0

1 + ( w

w0)2

Separating the transfer function in module and phase:

|G( jw)| =√

eR2 + X 2

|G( jw)| =

(H 0

1 + ( w

w0

)2)2 + (

H 0. w

w0

1 + ( w

w0

)2)2

φ( jw) = −tg−1(X

R)

φ( jw) = −tg−1(

H 0.w

w

01+( w

w0)2

H 0

1+( ww0

)2

)

φ( jw) = −tg−1(w

w0)

Bode Diagram

|H dB| = H 0dB + 20.log10.(1 + (w

wo

)2)−0.5

|H dB| = H 0dB − 10.log10.(1 + ( wwo

)2)

φ( jw) = −tg−1(w

w0)

For w << w0

|H dB| = H 0dB

φ( jw) = 0

For w = w0

|H dB| = H 0dB − 10log10(2)

φ( jw) = −π

4For w >> w0

|H dB| = H 0dB − 20log10(w

w0

)

φ( jw) = −π

2




Figure 3: Bode Diagram for Low-Pass Non-Inverter Filter (using H 0 = 10 and w0 = 500)

High-Pass Non-Inverter Filter

Figure 4: Circuit for High-Pass Non-Inverter Filter

The first part has a transfer function equal to:

V x

V in= G(s)1 =

s.C.R

1 + s.C.R

The second part has a transfer function as following:

V out

V x= G(s)2 =

Aol

1 + Aol.Z 2

Z 1+Z 2




Figure 5: Block Diagram for High-Pass Non-Iverter Filter

As Aol can ideally be considered infinity,

G(s)2 =Z 1 + Z 2

Z 2

The total transfer function would be:

G(s) = G(s)1.G(s)2 = (1 + Z 1

Z 2

).

s

s + 1CR

G(s) = H inf .s

s + w0

As s = σ + jw

and σ = 0

G( jw) = H inf .

jw

jw + w0 = H inf .

j

j + w0w

Separating the transfer function in module and phase:

|G( jw)| =√

eR2 + X 2

|H dB| = H inf dB + 20log10(w

w0

) − 10.log10(1 + (w

wo

)2)

φ( jw) = −tg−1(X

R)

φ( jw) = φ0 +π

2− tg−1(

w

w0)

Bode Diagram

|H dB| = H inf dB + 20log10(w

w0) − 10.log10(1 + (

w

wo

)2)

φ( jw) = φ0 +π

2− tg−1(

w

w0

)




For w << w0

|H dB

|= H inf dB + 20.log10(

w

w0

)

φ( jw) = φ0 +π

2

For w = w0

|H dB| = H inf dB − 10.log10(2)

φ( jw) = φ +π

4

For w >> w0

|H dB| = H inf dB

φ( jw) = φ0

Figure 6: Bode Diagram for High-Pass Non-Inverter Filter (using H inf = 10 and w0 = 500)

Conclusion

An analysis of non-inverter first order filters was successfully made. It was possible tocreate block diagrams and bode diagrams from the circuits and analyse the equations tofind amplitude and phase in function of frequency.


1st order filters-non inverters

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