exp2

experiment/procedure.md

@@ -18,7 +18,7 @@ L1-L8, L3-L4 or L4-L5, L3-L5, L6-L10 or L6-L9, L9-L10, L7-L13, L2-L12, L11-L12.(
 
 <div align="center">
 <img src="images/lowpass_prc.png" width="50%">
-<p>Figure 1</p>
+<p>Fig 1: Active Low Pass Filter</p>
 </div>
 
 - **Active High Pass Filter**
@@ -39,7 +39,7 @@ L1-L8, L3-L4 or L4-L5, L3-L5, L6-L10 or L6-L9, L9-L10, L7-L13, L2-L12, L11-L12.(
 
 <div align="center">
 <img src="images/highpass_prc.png" width="50%">
-<p>Figure 1</p>
+<p>Fig 2: Active High Pass Filter</p>
 </div>
 
 
@@ -63,5 +63,5 @@ L1-L10, L3-L12, L12-L14, L11-L13, L8-L13 or L8-L11, L4-L7, L4-L5 or L5-L7, L6-L9
 
 <div align="center">
 <img src="images/bandpass_prc.png" width="50%">
-<p>Figure 1</p>
+<p>Fig 3: Active Band Pass Filter</p>
 </div>

experiment/simulation/simulation/js/activebandpass.js

@@ -68,7 +68,7 @@ function tabled(){
 	var rld2=document.getElementById("rload2").value* Math.pow(10,3);	
 	var passbandgain1=(1+rf1/r1); 
 	var passbandgain2=(1+rf2/r2); 
-	var totalpassbandgain=passbandgain1+passbandgain2;
+	var passbandgain=passbandgain1+passbandgain2;
 	var flow= 1 / (2 * 3.14 *rld1 * parseFloat(cld1));
 	var fhigh= 1 / (2 * 3.14 *rld2 * parseFloat(cld2));
 

experiment/theory.md

@@ -29,7 +29,7 @@ $$ \frac{V_o}{V_i}= \frac{A_f}{\sqrt{1+(\frac{f}{f_h})^2}} $$
 
  $$ f_h=\frac{1}{2 \times \pi \times R_A \times C_A} $$ 
 
- Phase angle  (\(\phi\)) <br>
+ Phase angle  (ϕ) <br>
 &nbsp; $$ \phi = - atan(\frac{f}{f_h}) $$<br>
 
 We can choose the values of R<sub>f</sub> and R<sub>1</sub> suitably in order to obtain the desired gain at the output. Suppose, if we consider the resistance values of R<sub>f</sub> and R<sub>1</sub> as zero ohms and infinity ohms, then the above circuit will produce a unity gain low pass filter output.
@@ -44,13 +44,13 @@ The circuit diagram of an active high pass filter is shown in the following figu
 <p>Figure 2</p>
 </div>
 
-The electric network, which is connected to the non-inverting terminal of an op-amp is a passive high pass filter. So, the input of a non-inverting terminal of opamp is the output of passive high pass filter. The above circuit resembles a non-inverting amplifier. It is having the output of a passive high pass filter as an input to non-inverting terminal of op-amp. Hence, it produces an output, which is (1+\(\frac{R_f}{R_1}\))  times the input present at its non-inverting terminal.
+The electric network, which is connected to the non-inverting terminal of an op-amp is a passive high pass filter. So, the input of a non-inverting terminal of opamp is the output of passive high pass filter. The above circuit resembles a non-inverting amplifier. It is having the output of a passive high pass filter as an input to non-inverting terminal of op-amp. Hence, it produces an output, which is (1+R<sub>f</sub>/R<sub>1</sub>)  times the input present at its non-inverting terminal.
 
  &nbsp; $$ \frac{V_o}{V_i}= \frac{A_f \times \frac{f}{f_l}}{\sqrt{1+(\frac{f}{f_l})^2}} $$<br><br>
  where, $$ A_f=(1+\frac{R_f}{R_1}) $$ 
  $$ f_l=\frac{1}{2 \times \pi \times R_B \times C_B}$$ 
 
-Phase angle  $$ \phi $$<br>
+Phase angle  (ϕ)
 &nbsp;$$ \phi = atan(\frac{f}{f_l}) $$ <br>
 We can choose the values of R<sub>f</sub> and R<sub>1</sub> suitably in order to obtain the desired gain at the output. 
 Suppose, if we consider the resistance values of R<sub>f</sub> and R<sub>1</sub> as zero ohms and infinity ohms, then the above circuit will produce a unity gain high pass filter output. <br><br>