Linear and Nonlinear Elements
Вставка
- Опубліковано 29 сер 2018
- Network Theory: Linear and Nonlinear Elements
Topics discussed:
1) Linear elements
2) Law of homogeneity
3) Law of additivity
4) Nonlinear elements
5) Linear element example
6) Nonlinear element example
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Both are non-linear
->current becomes constant and does not increase with increase in voltage in 2nd problem
-> Non linear element must exhibit bidirectional property which is not followed in 4th problem
Correct sir! 4 one is symmetric along y axis but it need to be symmetric along origin .
This was an amazing video, you broke down and explained the concept so well. I just sat through a 1-hour lecture of my professor is has been trying to explain the exact same thing and it made absolutely no sense. I am mind blown that this 10-minute video explained everything so well.
Both are non linear as in second question as the decrease voltage current increases thus LOH is violated.
Also the line does not passes through origin.
Hats of man for this services 🙌
2- is non linear because after sometime when voltage is increased current becoms constant both side
4- is non linear because the straight line is not passing through origin
1:40 I would point out the i-v plot you showed was not for any resistor, but for a constant-resistance resistor, which we can prove it's a linear time-invariant device.
since there are two different lines are generating in 2 and 4 examples , it will have two different equations. and linear elements have only one equation . Hence both are non-linear
Bhai ,what is the final answer for the hw ,in the last case loh and loa are valid so according to me nl and l
Sir kindly give us some hits wether the Ans is correct or not which is given in Home work prblm daily
Thank you so much sir i m learning with conceptually each and every term clearly because of u u re great sir
Fig B is Non linear because of the constant and Fig D is Non Linear because the line is not passing through the origin .
Both are non linear
Nice concept
2-linear 'I' on constant increases as 'v' increases and back to constant after threshold.
4- non linear bcoz not passing through origin and also guess as v increases I drops to zerro.
It's passing through origin! But it's not symmetric about origin instead symmetric about y axis 🙂
2.Non leniar
4.Leniar
both are non linear coz they have equation of 2 st.lines.
Sir can you Tell me Linear cricuit / linear element both are same ????
2-nonlinear elements, 3-linear elements
How can I prove an *ideal* diode is also non-linear?
2) & 4) both non linear
Answer sir?
2 non linear
4 linear
are capictor and inductor are linear elements if yes, why ?
*Yes* if their inductance or capacitance is not a function of voltage nor current nor electric charge nor magnetic flux. The inductance or capacitance can be a function of time, and the inductor or capacitor are still linear, as long as the inductance or capacitance are not a function of voltage/current/charge/flux. For example, if L = 2 or C(t) = (log(t + 5) - e^t)/sin(t^2), that inductor and that capacitor are linear.
*Why yes?* Because differentiation and integration are linear operators.
*No* if their inductance or capacitance is a function of voltage and/or current and/or electric charge and/or magnetic flux. For example, if L(i(t),t) = i(t) - t or C(v(t),i(t)) = v(t)/i(t), that inductor and that capacitor are non-linear.
2nd is non linear and 4th is non linear because it does not follows one eqn. of st line
Both are non linear
Sir you aren't posting answers
both are non linear
Both Conundrums are Non-Linear. These Conundrums are not following LOH and LOA.
What the .........
Jenious duffer
Sir 2nd is non linear, this region is after some time the voltage is increase and current is constant or 4th is non linear element because the line passes through origin but it is not straight lines. Thank you sir
and one more thing is the line must me unique
Achha ok
Both are linear
2) liner
4) non liner
2nd nonlinear and 4th linear(because it's flow through the origin)
4th is also non-linear coz st.line is not passing through the origin
and also it has two eq's of a st.line
Linear element must exhibit bidirectional property which is not followed in 4th question
(Hw)
2. NONLINER
4.linear
The device of the 4th case is non-linear. Let's prove it.
Notice i(t) = |v(t)|. The device is linear if and only if it satisfies the superposition principle, which happens if and only if the device satisfies the homogeneity property and the additivity property.
Let's see if the device satisfies the homogeneity property.
First step. An input v_1(t) produces an output i_1(t) given by i_1(t) = |v_1(t)|.
Second step. We compute k · i_1(t): k · i_1(t) = k · |v_1(t)|.
Third step. An input v_2(t) given by v_2(t) = k · v_1(t) (where k is any constant real number) produces an output i_2(t) given by i_2(t) = |v_2(t)| = |k · v_1(t)|.
Fourth step. If i_2(t) = k · i_1(t) for all time t and for any real number k, the device satisfies the homogeneity property. Otherwise, the device does not satisfy this property. In our case, i_2(t) = |k · v_1(t)| is in general not the same as k · i_1(t) = k · |v_1(t)|. Therefore, the device does not satisfy the homogeneity property in general.
Let's see if the device satisfies the additivity property.
First step. An input v_3(t) produces an output i_3(t) given by i_3(t) = |v_3(t)|.
Second step. An input v_4(t) produces an output i_4(t) given by i_4(t) = |v_4(t)|.
Third step. We compute i_3(t) + i_4(t): i_3(t) + i_4(t) = |v_3(t)| + |v_4(t)|.
Fourth step. An input v_5(t) given by v_5(t) = v_3(t) + v_4(t) produces an output i_5(t) given by i_5(t) = |v_5(t)| = |v_3(t) + v_4(t)|.
Fifth step. If i_5(t) = i_3(t) + i_4(t) for all time t, the device satisfies the additivity property. Otherwise, the device does not satisfy this property. In our case, i_5(t) = |v_3(t) + v_4(t)| is in general not the same as i_3(t) + i_4(t) = |v_3(t)| + |v_4(t)|. Therefore, the device does not satisfy the additivity property in general.
Since the device does not satisfy both properties for all time t, for any constant real number k, and for any inputs, *the device is (in general) non-linear.*
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By the way, from properties of absolute value, we know that i_2(t) = |k · v_1(t)| = |k| ·| v_1(t)|. If we assume k >= 0, then i_2(t) = k ·| v_1(t)|, which is the same as k · i_1(t) = k · |v_1(t)|. In other words, while the device in general does not satisfy the homogeneity property, it does satisfy it if k is non-negative.
By the way, from properties of absolute value, we know that i_5(t) = |v_3(t) + v_4(t)| 0 and v_4 > 0, or, v_3 < 0 and v_4 < 0), then i_5(t) = |v_3(t)| + |v_4(t)|, which is the same as i_3(t) + i_4(t) = |v_3(t)| + |v_4(t)|. In other words, while the device in general does not satisfy the additivity property, it does satisfy it if v_3 and v_4 have the same sign.
So, if we restrict that any scaling factor must be real non-negative, and that when summing two inputs both must have the same sign, then the device satisfies both properties and so *the device is linear in that particular scenario.*
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9:07 Usually, "characteristics curve" refers to the curve given by plotting i(t) in one axis and v(t) in another axis in a plane. I will assume this is what you meant.
But then, it is a false claim that a linear device will have a characteristics curve given by a straight line through the origin. It's true *all* devices with that curve are linear, but it's not true *all* linear devices have that curve.
As an example, consider a constant-inductance inductor. This is of course a linear (and time-invariant) device. Now suppose we apply a sinusoidal voltage, resulting in a sinusoidal current of same frequency. The characteristics curve (current-voltage plot) of such an inductor will be an ellipse, not a straight line through the origin!
Both are non linear