Page 1 of 16 ECEN 622 Homework #1

Texas A&M University Department of Electrical and Computer Engineering

ECEN 622: Active Network Synthesis Homework #1, Fall 2016

Carlos Pech Catzim 723002156

Page 2 of 16 ECEN 622 Homework #1

1. Design a Low Pass filter with the following specs: HLP(0) = 1, wo= 2π x 106 rad/s, Q = 2. Using the Sallen and Key unity gain implementation.

a) Assume an ideal Op Amp and determine the component values and plot the magnitude and phase vs frequency.

R2

C1

C2

R1

VoVi

Figure 1. Sallen and Key Unity Gain Topology.

Assuming and ideal operational amplifier we can use nodal analysis to obtain the following transfer function.

𝐻𝐿𝑃(𝑠) =

1𝑅2𝐶1𝐶2

𝑠2 + 𝑠 (2

𝑅𝐶2) + (

1𝑅2𝐶1𝐶2

)=

𝑤𝑜2

𝑠2 +𝑤𝑜

𝑄𝑠 + 𝑤𝑜

2

(1)

Where we assumed that 𝑅1 = 𝑅2 = 𝑅, 𝐶1 = 𝐶, 𝐶2 = 4𝐶𝑄2, and 𝑅𝐶 =1

2𝑤𝑜𝑄. Using the previous assumptions

and using the desired values for gain and cut-off frequency we can find the values for our components.

COMPONENT VALUE

R 40k Ω C1 1p F C2 16p F

Figure 2. Sallen and Key Unity Gain Topology frequency response with Ideal Op Amp.

Page 3 of 16 ECEN 622 Homework #1

Figure 2 shows the behavior of our filter, the magnitude stays close to 0dB with peaking of 6dB at 𝑤𝑜 due

to Q = 2, as 𝐻(𝑗𝑤𝑜) ≈ 𝑄. The phase also shows the expected behavior of a second order low pass filter with ≈ -90° of phase at 𝑤𝑜 and -180° of phase at higher frequencies.

b) Assume now a real Op amp characterized by: 𝑨(𝒔) =𝑨𝒐

𝟏+𝒔/𝒘𝒑=

𝑮𝑩

𝒔+𝒘𝒑, with macromodel shown below.

Let 𝒘𝒑 = 𝟐𝝅𝒙𝟏𝟎𝟑𝒓𝒂𝒅/𝒔 Determine the value of GB yielding a magnitude or phase error of less than

1% of the filter.

Figure 3. Non Ideal Op Amp Macromodel

From the equation given we can find that 𝑤𝑝 = 1/𝑅𝑝𝐶𝑝 which then we can use to find the desired values.

COMPONENT VALUE

RP 100kΩ CP 1.6nF

Now in order to find the value of GB we need to find a transfer function which takes the non-ideality of the opamp into account, resulting in the following expression.

𝐻(𝑠) =

1

(1 +1

𝐴(𝑠)) 𝑅2𝐶1𝐶2

𝑠2 + 𝑠 [1

𝑅𝐶1+

2𝑅𝐶2

− (1

𝑅𝐶1 (1 +1

𝐴(𝑠))

) ] +1

𝑅2𝐶1𝐶2

(2)

Now using the previously found values for R, C1 and C2, 𝑤𝑜 = 1/4𝑅𝐶, in addition to the approximation

1

1+1/𝐴(𝑠)≈ 1 − 1/𝐴(𝑠) for |𝐴(𝑠)| ≫ 1, we can derive an expression for the error, the new Q and new 𝑤𝑜.

𝐻(𝑠) =𝑤𝑜

2 (1 −1

𝐴(𝑠))

𝑠2 + 𝑠𝑤𝑜 (0.5 +4

𝐴(𝑠)) + 𝑤𝑜

2=

𝑤𝑜2 (1 −

𝑠 + 𝑤𝑝

𝐺𝐵)

𝑠2 + 𝑠𝑤𝑜 (0.5 +4(𝑠 + 𝑤𝑝)

𝐺𝐵) + 𝑤𝑜

2

𝐻(𝑠) = 𝑤𝑜

2 (1 −𝑠 + 𝑤𝑝

𝐺𝐵)

𝑠2(1 +4𝑤𝑜

𝐺𝐵) + 𝑠𝑤𝑜 (0.5 +

4𝑤𝑝

𝐺𝐵) + 𝑤𝑜

2

Where we can define 1 + 𝑒 = 1 + 4𝑤𝑜/𝐺𝐵 or 𝑒 =4𝑤𝑜

𝐺𝐵≈

4

|𝐴(𝑗𝑤𝑜)|

Now we redistribute H(s) as

𝐻(𝑠) =

𝑤𝑜2

1 + 𝑒(1 −

𝑠 + 𝑤𝑝

𝐺𝐵)

𝑠2 + 𝑠 [𝑤𝑜

√1 + 𝑒

(0.5 +4𝑤𝑝

𝐺𝐵)

√1 + 𝑒] +

𝑤𝑜2

1 + 𝑒

Page 4 of 16 ECEN 622 Homework #1

Now we have that

𝑤𝑜𝑅 =𝑤𝑜

√1 + 𝑒≈ 𝑤𝑜 (1 −

𝑒

2) ≈ 𝑤𝑜 − Δ𝑤𝑜 , 𝑤ℎ𝑒𝑟𝑒 Δ𝑤𝑜 =

2𝑤𝑜

𝐺𝐵

𝑄𝑅 =√1 + 𝑒

0.5 +4𝑤𝑝

𝐺𝐵

≈1 + 𝑒/2

0.5 +4𝑤𝑝

𝐺𝐵

ℎ = 1 −𝑠 + 𝑤𝑝

𝐺𝐵

In the previous expressions we can see how the non-idealities of the op amp affects wo, Q and creates a zero.

In order to determine the value of GB yielding a magnitude error < 1% we calculate the magnitude of (1) and (2) using MATLAB and we find that the necessary GB = (2π) (72.5177 MHz), or Ao = 72.5177k V/V = 97.2dB. Figure 4 shows the response of the designed non-ideal op amp.

Figure 4. Non-ideal Op Amp.

Figure 5. Sallen and Key Unity Gain Topology frequency response with non-Ideal Op Amp.

Page 5 of 16 ECEN 622 Homework #1

Figure 6. Sallen and Key Unity Gain Topology frequency response Comparison.

In figure 5 we show the frequency response of the filter using the non-ideal op amp, and in figure 6 we show a comparison of the two with markers that show that both responses behave within the 1% of error within the band of operation.

The calculations made demand for an extremely high open loop gain, however to achieve such gain we would have to design a three or four stages op amp, that comes with the challenge of compensation and power.

Measurement Ideal Non-Ideal Error

Q 2 2.009 -0.45 %

𝑤𝑜 2π 106 2π 972.42k -2.83%

2. Design for the same LP specs a KHN biquad

R3

R1

R2

R5

R4

R6 R7C1 C2

Vi

Vo

Figure 7. KHN Biquad Schematic.

Page 6 of 16 ECEN 622 Homework #1

a) Determine the conditions of R1, R2, R3, R4, and R5 such that the adder is fully balanced, then determine the component values of the KHN biquad.

As discussed in class in order for the summing amplifier to be fully balanced we need that the sum of the admittances at the input nodes of the op amp to be equal, that is:

1

𝑅3+

1

𝑅4+

1

𝑅5=

1

𝑅1+

1

𝑅2= 𝑌𝑇𝑜𝑡𝑎𝑙

To better represent the system, and simplify the calculation of the transfer function along with the case of non-ideality we draw the block diagram of the circuit as shown in figure 8.

Figure 8. Block Diagram Representation of KHN biquad with ideal opamps

Where the gain values are

𝐾1 = −𝑅5

𝑅3 , 𝐾2 = −

1

𝐶1𝑅6, 𝐾3 = −

𝑅5

𝑅4, 𝐾4 =

𝑅5

𝑅2, 𝐾5 = −

1

𝐶2𝑅7

Now using mason rules, we can find the transfer function

𝐻(𝑠) = −𝐻𝑜𝑤𝑜

𝑠2 + 𝑠 (𝑤𝑜

𝑄) + 1

= −𝐾1/𝐾3(𝐾2𝐾5𝐾3)

𝑠2 + 𝑠(𝐾2𝐾4) + 𝐾2𝐾5𝐾3

𝐻𝑜 represents an extra gain coefficient which forces 𝐾1 = 𝐾3 𝑜𝑟𝑅4

𝑅3= 1, furthermore we can express 𝑄 =

1/𝐾4 which points to a condition of 𝑅2 = 2 𝑅5 and 𝑤𝑜 = 𝐾2 = 1/𝐶1𝑅6, in order to fulfill the condition for a fully balanced circuit we will make 𝑅3 = 𝑅4 = 𝑅5, and 𝑅3 = 2.5𝑅1, now we have an expression for every

component except for R7 which will be calculated using the relationship 𝑅7 =1

𝑤𝑜2𝐶1𝑅6𝐶2

.

Component Value Component Value

R3, R4, R5 10KΩ R2 20KΩ

R1 4KΩ C1 1pF

R6 160kΩ C2 1.5pF R7 105kΩ

b) Using Simulink simulate your designed filter for

i) Ideal Op Amp

Using the same block diagram from figure 8 we are able to simulate the frequency response of the KHN filter as shown in figure 9. Again we notice the peaking at wo due to the value of Q, also we notice that this time the phase starts at -180° as this configuration is provides an inverting low pass behavior.

Page 7 of 16 ECEN 622 Homework #1

Figure 9. Frequency Response for KHN Filter with ideal Op Amp

ii) A non-ideal Op Amp with A(s) ≈GB/s when GB = 20wo.

For this case we have to consider the non-ideality of each op amp used in the circuit. First we will start with the summing amplifier.

Previously we used the property of sum of admittances so the gain of each input would be given by a ratio of resistors, this time as the voltage at both inputs is not equal we must rewrite the node equations using admittances as:

R3

R1

R2

R5

R4

Vo

V2

Vi

V1

Figure 10. Non Ideal Op Amp, Summing Amplifier.

𝑉−(𝑌3 + 𝑌4 + 𝑌5) − 𝑉𝑖𝑌3 − 𝑉2𝑌4 = 𝑉𝑜𝑌5 (3) 𝑉+(𝑌1 + 𝑌2) − 𝑉1𝑌2 = 0 (4)

𝐴(𝑠)(𝑉+ − 𝑉−) = 𝑉𝑜 (5)

Page 8 of 16 ECEN 622 Homework #1

Now we make 𝑌𝑡𝑜𝑡𝑎𝑙 = 𝑌1 + 𝑌2 = 𝑌3 + 𝑌4 + 𝑌5, use (5) and substract (4) of (3) we obtain:

𝑉𝑜 =1

𝑌5 +𝑌𝑇𝑜𝑡𝑎𝑙

𝐴(𝑠)

[𝑉1𝑌2 − 𝑉𝑖𝑌3 − 𝑉2𝑌4] =1

1 +𝑍5𝑌𝑇𝑜𝑡𝑎𝑙

𝐺𝐵/𝑠

[𝑉1𝑍5/𝑍2 − 𝑉𝑖𝑍5/𝑍3 − 𝑉2𝑍4/𝑍5]

As before the gain from each input is given by the ratio of impedances but now there is a scaling factor for each input which depends on GB and the frequency.

For the lossless integrators we have that instead of using the ideal integrator block 1/𝑠 we will use a block that represents the losses due to the non-ideality of the op amp. This case in particular was reviewed in class where we found that the new expression for the lossless integrator is given by

𝐻𝑖𝑛𝑡(𝑠) = −1

𝑠𝑅𝐶 (1 +𝑠

𝐺𝐵)

Where R and C are the impedances connected to the operational amplifier. Now use the block diagram shown in figure 11 and simulate the frequency response shown in figure 12.

Figure 11. Block Diagram Representation of KHN filter with non-ideal Op Amp

Figure 12. Frequency Response for KHN Filter with non-ideal Op Amp.

Page 9 of 16 ECEN 622 Homework #1

We notice that using GB = 20GB causes the gain peaking around wo doubled as each individual non-ideal integrator introduces an error magnitude and phase of approximately 2.5%, and the non-ideal summing amplifier introduces another error. In the next figure we show that increasing the value of GB minimizes the error.

Figure 13. Frequency Response for KHN Filter with non-ideal Op Amp, multiples GB.

Similar with the previous exercise the non-idealities of the op amp causes error in the desired frequency response, that can only be fix increasing the open loop gain of the op amp or to change design parameters accordingly.

Figure 14. Group Delay for KHN Filter with ideal and non-ideal Op Amp.

Page 10 of 16 ECEN 622 Homework #1

Measurement Ideal Op Amp Non-Ideal Op Amp

(GB=20wo) Error %

(GB=20wo) Non-Ideal Op Amp

(GB=50wo) Error %

(GB=50wo)

Magnitude @ wo 6dB 12dB 99% 7.98dB 25.3%

Phase @ wo -269° -271° 0.743% -270° 0.371.75%

Group Delay @ wo 84.47ns 81.5ns -3.5% 83.29ns -1.397%

3. Design a RC Miller integrator for 𝒘𝒐 = 𝟐𝝅𝒙𝟏𝟎𝟔𝒓𝒂𝒅/𝒔

a) Assume an ideal Op Amp.

R C

Vi Vo

Figure 15. Miller Integrator

We know that the miller integrator has a transfer function 𝐻(𝑠) =1

𝑠𝑅𝐶, where we can easily find the

component values by setting a desired C and finding 𝑅 =1

𝑤𝑜𝐶.

Component Value Component Value

R 31.8kΩ C 5pF

Figure 16. RC Miller Integrator Frequency Response, Ideal Op Amp.

Page 11 of 16 ECEN 622 Homework #1

In figure 16 we show that the Miller integrator follows the expected frequency response.

b) Let 𝑮𝑩 = 𝟐𝝅𝒙𝟓𝒙𝟏𝟎𝟔 for 𝑨(𝒔) = 𝑮𝑩/𝒔

Assuming a non-ideal integrator, we have the following transfer function

𝐻𝑛𝑜𝑛(𝑠) =1

𝑠𝐶𝑅 (1 +𝑠

𝐺𝐵)

Figure 17. RC Miller Integrator Frequency Response, Non-Ideal Op Amp.

In figure 17 we show that the non-ideal op amp introduces an extra pole that creates an error in phase given by

𝜃(𝑤) = −90° − tan−1(𝑤/𝐺𝐵) ≈ 101.3°

Δ𝜃(𝑤) =90 − 101.3

90≈ 12.5%

c) Add a resistor 𝑹𝒄 in series with the feedback capacitor such that an ideal integrator is obtained.

By placing a 𝑅𝑐 resistor in series with the feedback capacitor we create a zero that can cancel the parasitic

pole created by the op am, given that the resistor value is well defined. This cancellation occurs if 𝑅𝑐 =1

𝐺𝐵𝐶= 6.36𝑘Ω.

R C

Vi

Vo

Rc

Figure 18. Compensated RC Miller Integrator

Page 12 of 16 ECEN 622 Homework #1

𝐻(𝑠) =1 + 𝑅𝑐𝐶𝑠

𝑅𝐶𝑠(

1

1 +𝑠

𝐺𝐵+

1𝐺𝐵𝑅𝐶

+𝑠𝑅𝑐

𝐺𝐵𝑅

) =1 + 𝑠/𝐺𝐵

𝑅𝐶𝑠(

1

1 +1

𝐺𝐵𝑅𝐶+

𝑠𝐺𝐵

(1 +1

𝐺𝐵𝑅𝐶)

)

= 1 + 𝑠/𝐺𝐵

𝑅𝐶𝑠(

1

(1 +𝑠

𝐺𝐵) (1 +

1𝐺𝐵𝑅𝐶

) ) =

1

𝑅𝐶𝑠(

1

(1 +1

𝐺𝐵𝑅𝐶)

) =1

𝑅𝐶𝑠 (1 +1

𝐺𝐵𝑅𝐶)

From the equation above we can see that our new transfer function 𝐻(𝑠) only has imaginary part which corrects the error in phase but introduces an error in magnitude given by

|𝐻(𝑗𝑤𝑜)| =1

𝑅𝐶𝑤𝑜 +𝑤𝑂

𝐺𝐵

≈ 833.33𝑚𝑉 ≈ −1.5836𝑑𝐵

Figure 19. RC Miller Integrator Frequency Response, Compensated Non-Ideal Op Amp.

Measurement Ideal Op Amp Non-Ideal Op Amp Compensated Op Amp

|𝑯(𝒋𝒘𝒐)| -0.02267dB -0.177dB -1.59dB

𝜽(𝒋𝒘𝒐) -90° -101° -90°

Page 13 of 16 ECEN 622 Homework #1

Extra Credit, Repeat problem 1 for the LP Rauch Filter, for HLP(0) = 1, wo = 2π x106 rads/s, Q = 2.

a) Assume an ideal Op Amp and determine the component values and plot the magnitude and phase vs frequency

R3

C1

C2

R2

VoVi

R1

Figure E1. Rauch Low Pass Filter

Assuming an ideal operational amplifier, we can use nodal analysis to obtain the following transfer function

𝐻𝐿𝑃(𝑠) =

− (𝑅1

𝑅3) (

1𝑅1𝑅2𝐶1𝐶2

)

𝑠2 + 𝑠 (𝑅1𝐶2 + 𝑅2𝐶2 + 𝑅1𝑅2𝐶2/𝑅3

𝑅1𝑅2𝐶1𝐶2) + (

1𝑅1𝑅2𝐶1𝐶2

)=

𝐻𝑜𝑤𝑜2

𝑠2 +𝑤𝑜

𝑄𝑠 + 𝑤𝑜

2

(E1)

Where 𝑤𝑜 = 1/√𝑅1𝑅2𝐶1𝐶2 , 𝑄 = √𝑅1𝑅2𝐶1𝐶2

𝑅1𝐶2+𝑅2𝐶2+𝑅1𝑅2𝐶2/𝑅3, 𝐻𝑜 = 𝑅1/𝑅3. Now Assuming 𝑅 = 𝑅1 = 𝑅2 = 𝑅3, 𝐶 = 𝐶2,

and 𝐶1 = 𝑛𝐶 we can rewrite our parameters as

𝑤𝑜 =1

𝑅𝐶√𝑛 𝑎𝑛𝑑 𝑄 =

√𝑛

3

Using the previous assumptions and using the desired values for gain and cut-off frequency we can find the values for our components.

COMPONENT VALUE

R 10.2k Ω C1 93.6p F C2 2.6p F

Figure E2. LP Rauch Filter Frequency Response with Ideal Op Amp.

Page 14 of 16 ECEN 622 Homework #1

Figure E2 shows the expected behavior of the filters frequency response, as happened in the salley and key filter the peaking at wo appears due to the value of Q.

b) Assume now a real Op amp characterized by: 𝑨(𝒔) =𝑨𝒐

𝟏+𝒔/𝒘𝒑=

𝑮𝑩

𝒔+𝒘𝒑, with macromodel shown below.

Let 𝒘𝒑 = 𝟐𝝅𝒙𝟏𝟎𝟑𝒓𝒂𝒅/𝒔 Determine the value of GB yielding a magnitude or phase error of less than

1% of the filter.

Figure E3. Non Ideal Op Amp Macromodel

For the non-ideal op amp, we will use the same that was designed and used in previous exercises.

Now in order to find the value of GB we need to find a transfer function which takes the non-ideality of the opamp into account, resulting in the following expression.

𝐻(𝑠) =

𝐻𝑜𝑊𝑜2 1

(1 +1

𝐴(𝑠))

𝑠2 + 𝑠 [𝑤𝑜

𝑄+

𝑤𝑜2𝑅1𝐶1

1 + 𝐴(𝑠) ] + 𝑊𝑜

2 [1 +

𝑅1

𝑅3

1 + 𝐴(𝑠) ]

(E2)

Now using the approximation 1

1+1/𝐴(𝑠)≈ 1 − 1/𝐴(𝑠) for |𝐴(𝑠)| ≫ 1, we can derive an expression for the

error, the new Q and new 𝑤𝑜.

𝐻(𝑠) =𝐻𝑜𝑊𝑜

2 (1 −1

𝐴(𝑠))

𝑠2 + 𝑠 [𝑤𝑜

𝑄+

𝑤𝑜2𝑅1𝐶1

1 + 𝐴(𝑠) ] + 𝑊𝑜

2 [1 +

𝑅1

𝑅3

1 + 𝐴(𝑠) ]

𝐻(𝑠) =𝐻𝑜𝑊𝑜

2 (1 −𝑠 + 𝑤𝑝

𝐺𝐵)

𝑠2 + 𝑠 [𝑤𝑜

𝑄+

𝑤𝑜2𝑅1𝐶1(𝑠 + 𝑤𝑝)

𝐺𝐵 ] + 𝑊𝑜

2 [1 +

𝑅1

𝑅3(𝑠 + 𝑤𝑝)

𝐺𝐵 ]

𝐻(𝑠) =𝐻𝑜𝑊𝑜

2 (1 −𝑠 + 𝑤𝑝

𝐺𝐵)

𝑠2[1 +𝑅1𝐶1𝑤𝑜

2

𝐺𝐵] + 𝑠 [

𝑤𝑜

𝑄+

𝑤𝑜2𝑅1𝐶1𝑤𝑝

𝐺𝐵+

𝑤𝑜2𝑅1/𝑅3

𝐺𝐵 ] + 𝑊𝑜

2 [1 + 𝑤𝑝𝑅1/𝑅3

𝐺𝐵 ]

Where we can define 1 + 𝑒 = 1 + 𝑅1𝐶1𝑤𝑜2/𝐺𝐵 or 𝑒 =

6𝑤𝑜

𝐺𝐵≈

6

|𝐴(𝑗𝑤𝑜)|

Page 15 of 16 ECEN 622 Homework #1

Now we redistribute H(s) as

𝐻(𝑠) =𝐻𝑜𝑊𝑜

2 (1 −𝑠 + 𝑤𝑝

𝐺𝐵) (

11 + 𝑒

)

𝑠2 + 𝑠 [𝑤𝑜

√1 + 𝑒] [

1𝑄

+𝑤𝑜𝑅1𝐶1𝑤𝑝

𝐺𝐵+

𝑤𝑜𝑅1/𝑅3

𝐺𝐵

√1 + 𝑒 ] + 𝑊𝑜

2 [1 +

𝑤𝑝𝑅1/𝑅3

𝐺𝐵 1 + 𝑒

]

Now we have that

𝑤𝑜𝑅 =𝑤𝑜

√1 + 𝑒≈ 𝑤𝑜 (1 −

𝑒

2) ≈ 𝑤𝑜 − Δ𝑤𝑜 , 𝑤ℎ𝑒𝑟𝑒 Δ𝑤𝑜 =

6𝑤𝑜

𝐺𝐵

𝑄𝑅 =√1 + 𝑒

1𝑄

+𝑤𝑜𝑅1𝐶1𝑤𝑝

𝐺𝐵+

𝑤𝑜𝑅1/𝑅3

𝐺𝐵

≈1 + 𝑒/2

1𝑄

+𝑤𝑜𝑅1𝐶1𝑤𝑝

𝐺𝐵+

𝑤𝑜𝑅1/𝑅3

𝐺𝐵

𝑤𝑧 = 1 −𝑠 + 𝑤𝑝

𝐺𝐵

In the previous expressions we can see how the non-idealities of the op amp affects wo, Q and creates a zero, without affecting the DC gain.

In order to determine the value of GB yielding a magnitude error < 1% we calculate the magnitude of (1) and (2) using MATLAB and we find that the necessary GB = (2π) (232.4MHz), or Ao = 232.4k V/V = 107.32dB.

Figure E4. LP Rauch Filter frequency response with non-Ideal Op Amp.

Page 16 of 16 ECEN 622 Homework #1

Figure E5. LP Rauch Filter frequency response Comparison.

In figure E4 we show the frequency response of the filter using the non-ideal op amp, and in figure E5 we show a comparison of the two with markers that show that both responses behave within the 1% of error in magnitude within the band of operation, for a 1% error in phase a much higher open loop gain is needed

The calculations made demand for an extremely high open loop gain, however to achieve such gain we would have to design a three or four stages op amp, that comes with the challenge of compensation and power.

Measurement Ideal Non-Ideal Error

Q 2 1.98 -1 %

𝑤𝑜 2π 106 2π 987.09k -1.3 %

An extra table is provided in which we show the comparison between the Sallen and key, and KHN low-pass filters with the same specs considering non-ideal op amps.

Measurement Sallen and Key KHN Rauch Ideal

Magnitude @ wo 5.97085dB 6.1dB 5.934dB 6dB Phase @ wo -91.7° -91.7° 86.85887° -90°

Minimum GB for 1% error

72.51 * wo 1200 * wo 232.4 * wo -

The minimum GB needed in the KHN low pass filter is higher due to the accumulative error in each stage. We could reduce the GB needed by Q-compensation schemes in order to reduce the overall error of the system.