Excercises
==========



   
Equivalent Circuits
-------------------

We have already seen that two resistors in series can be replaced with
one resistor. Thus the circuit

    .. tikz::

       \draw (0,0) to[R=\$R_1\$,o-] (2,0) to[R=\$R_2\$,-o] (4,0);

is equivalent with one resistor

    .. tikz::

       \draw (0,0) to[R=\$R_1+R_2\$,o-o] (4,0);

We also have that the parallel connection of two resistors:

    .. tikz::

       \draw (0,0)
       to[short,o-] (1,0)
       to[short] (1,0.6)
       to[R=\$R_1\$] (3,0.6)
       to[short] (3,0)
       to[short,-o] (4,0);
       \draw (1,0)
       to[short] (1,-0.6)
       to[R=\$R_2\$] (3,-0.6)
       to[short] (3,0);

is equivalent with one resistor:

    .. tikz::

       \draw (0,0) to[R=\$\frac{R_1 R_2}{R_1+R_2}\$,o-o] (4,0);


#. The entire circuit below can be replaced with
   just one resistor, what is the value of this resistor?

   .. tikz::

      \draw (0,3)
      to[R=\$R_1\$,o-] (4,3)
      to[R=\$R_3\$] (8,3)
      to[R=\$R_4\$] (8,0)
      to[short,-o] (0,0);
      \draw (4,3) to[R=\$R_2\$] (4,0);


#. The same applies for the circuit below. Surpringly enough the
   simple series and parallel resistors tricks won't help you out
   here.

   .. tikz::

      \draw (0,0) to[short,o-] (1,0);
      \draw (1,-1) 
      to[short] (1,1)
      to[R=\$R_1\$] (3,1)
      to[R=\$R_2\$] (5,1)
      to[short] (5,-1);
      \draw (5,0) to[short,-o] (6,0);
      \draw (1,-1) 
      to[R=\$R_3\$] (3,-1)
      to[R=\$R_4\$] (5,-1);
      \draw (3,1) to[R=\$R_5\$] (3,-1);

   The use of the Kirchhoff laws will help you out here. Or look on
   the web for the $Y-\Delta$ transformation.


Complex Impedance
-----------------

If you read $Z, Z_1$ etc. instead of $R, R_1$ etc. in the equivalent
circuits in the previous exercise then the results are perfectly
useable for inductors, capacitors and combinations with resistors.

#. Show that the series connection of two inductors with inductances
   of $L_1$ and $L_2$ Henry is equivalent with an inductor of $L_1+L_2$
   Henry.

#. What is the equivalent inductance $L$ of two inductors ($L_1$ and
   $L_2$) connected in parallel?

#. What is the equivalent capacity of two capacitors connected in
   series?

#. What is the equivalent capacity of two capacitors connected in
   parallel?

       
Control Sound Level
-------------------

The amplifier in your stereo system amplifies the analog signal with a
small amplitude to a signal of larger amplitude and more power to
drive your speakers.
		 
   .. tikz::

      \input{/home/rein/Dropbox/UvA/SP/DSPNOTES/components}
      \draw (0,4)
      to[short,o-] (2,4)
      to[R=\$R_1\$] (2,2)
      to[R=\$R_2\$] (2,0)
      to[short,-o] (0,0);
      \draw (2,2)
      to[short] (4,2)
      %to[generic,n=S1] (4,0)
      to[R, l=Speaker (\$8\Omega\$)] (4,0)
      to[short] (2,0);
      %\speaker{S1}{0}


At home i have a set of speakers connected to the amplifier i the
living room and a second set of speakers in the kitchen connected to
the same amplifier. Works great but unfortunately the speakers in the
kitchen need less energy (lower amplitude) to generate the same sound
level as the speakers in the living room. So either the music in the
kitchen is too loud or the music in the living room is too soft.

To make sound levels equal in both areas we can use a simple *voltage
divider* as depicted here. The voltage $u_1$ is from a source (the
amplifier) measured at the input terminals on the left.

Calculate the voltage $u_2$ (over the speaker) in terms of $u_1$,
$R_1$ and $R_2$. The speaker itself can be seen as a resistor and
should be taken into account.

Keep in mind that the amplifier is build to expect a *load* of about
$8 \Omega$. How would you choose $R_1$ and $R_2$ such that $u_2 =
u_1/3$ *and* the load for the amplifier is still about $8 \Omega$.

.. 
   #. If done correctly the circuit you designed will work correctly
   (please use resistors that are made for the amount of current and
   thus power you can expect... else you get a result that is too
   hot). But replacing the two resistors with a *potentiometer*
   enables you to control the sound level in the kitchen from no sound
   to full sound.

   Look up on the web what a potentiometer is and how to use it to
   replace the two resistors in the circuit. Again be sure to select a
   potmeter (you might as well learn the slang) that is capable of the
   power you throw at it. High power potmeters are nog cheap! 

   The potmeter in your amplifier is used in more or less the same
   voltage dividing function. But it is done for the signal that is
   not amplified yet, this makes that the potmeter doesn't have to
   cope with a lot of power (and thus can be cheaper).


Design 2-way X-over
-------------------

#. Now consider the task of designing a cross-over filter for a two
   way speaker system. Both drivers, the woofer and the tweeter, are 8
   $\Omega$. Again a cross-over frequency of around 2 kHz is
   required. We will use the simple filters as discussed in previous
   sections. Plot in one figure the magnitude of the frequency
   response of the low-pass $H_L(\omega)$ and high-pass $H_H(\omega)$
   filters.

   Also plot the magnitude of their sum (note: first adding then
   taking the magnitude). In case of ideal speakers (none of them are
   in reality of course) you want a perfect constant total frequency
   response. Is that what you get?

#. At this `webpage <http://www.diyaudioandvideo.com/Calculator/XOver/>`_
   you find a calculator for two way x-over design. You can check your
   design using the calculator and setting the crossover type to "1st
   order Butterworth" (not quite as in a Butterworth filter the corner
   frequencies for the low and high pass filter are not precisely
   equal but close enough in practice).

   Use the calculator to design a 1st order filter using the "Solen
   Split" setting. What are the corner frequencies for the high and
   low pass filters in this case?

   .. figure:: /figures/philips_xover.*
      :align: right
      :figwidth: 40%

#. Let's return to the Philips x-over filter. Their high-pass filter
   is not quite what we have done before. They have added an inductor
   parallel to the driver.

   Calculate the frequency response of the high pass filter (assuming
   an inductance $L$ and capacitance $C$ in the high pass
   filter). Make a Bode plot of the magnitude. What is the corner
   frequency and what is the decay in dB per octave?




RIAA Correction
---------------

I still play vinyl records that i collected in the days CD's were non
existing (and no, there was no streaming music in those days because
there was no digital music and there was no internet ...).

As you know vinyl records store music in an analog form. The grooves
in the record encode the changing sound pressure. The needle follows
the groove and will start to vibrate. This vibration (movement) of the
needle makes a magnet move with respect to a coil (or the other way
round if you have a deep pocket) and induce a voltage that is
propertional to the sound level.

We won't go into any of the nice details like:

- how do you encode the left and right sound signals using just one
  needle (see nice `animation
  <http://www.vinylrecorder.com/stereo.html>`_), and

- how do you get the grooves in the record in the first place, and

- how do you deal with the fact that the record is spinning at the
  same speed and thus the length of groove the needle travels in one
  second depends on whether the needle is at the start or the end of
  the record,

and other nasty practical problems. One problem we do look it in this
exercise is the fact that low frequencies tend to be close to the
mechanical eigen vibrations of the needle system. To much low
frequencies would throw the needle out of the groove. Therefore the
RIAA system was developed. Before the audio signal is transformed into
grooves on the record the signal is filtered. Low frequencies are
attenuated. But then when the needle follows the groove and you would
not correct the sound signal you would hear a sound with too much high
frequency content. So we have increase the low frequencies and
decrease the high frequencies. There are a lot of ways to make such a
filter. Here we discuss a simple (very simple..) analog filter using a
few passive components. We only give a mono version.

..   Here explanation of:
     - constant angular velocity (RPM)
     - varying velocity along track
     - compare with CD
     - rumble, flutter and all that (mechanical problems...)
     - why RIAA is used

In order to restore the original sound signal we need to filter the
signal coming from the record player using the RIAA decoding
curve. A passive filter to do this is sketched:

.. tikz::

   \draw (0,5)
   to[R=\$75k\Omega\$,o-] (2,5)
   to[short,-o] (4,5);
   \draw (2,5)
   to[C=\$0.02916\mu F\$] (2,3)
   to[R=\$10.905 k\Omega\$] (2,1)
   to[short] (3,1);
   \draw (3,5)
   to[C=\$0.01\mu F\$] (3,0) node[tground] {};
   
.. this is comment, in fact this is the encoding riaa filtering
   
   .. tikz::

      \draw (0,0)
      to[short,o-,l=\$u_i\$] (1,0)
      to[R] (3,0)
      to[R] (5,0)
      to[short,-o, l=\$u_o\$] (6,0);
      \draw (1, 0)
      to[short] (1, 1)
      to[C] (3, 1)
      to[C] (5, 1)
      to[short] (5, 0)
      to[R] (5,-2) node[tground] {};
      \draw (3,0)
      to[short] (3,1);
   
#. Calculate the complex transfer function

#. Plot the frequency response of this passive filter (both amplitude
   as well as phase).