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# Circuit Theory/Convolution Integral/Examples/example49/VL

Given that the source voltage is (2t-3t$^2$), find voltage across the resistor.
This is the V$_L$ solution.
Outline:

### Transfer Function

    $H(s)={\frac {V_{L}}{V_{S}}}={\frac {s}{4+s+{\frac {1}{0.25s}}}}$
    $H(s)={\frac {s^{2}}{s^{2}+4s+4}}$

### Homogeneous Solution

There are two equal roots at s = -2, so the solution has the form:
    $V_{L_{h}}(t)=Ae^{-2t}+Bte^{-2t}+C_{1}$

### Particular Solution

After a long time attached to a unit step function source, the inductor has shorted and the capacitor has opened. All the drop is across the capacitor.
    $V_{L_{p}}=0$
This also means that C$_1$ has to be zero.

### Initial Conditions

So far the full equation is:
    $V_{L}(t)=Ae^{-2t}+Bte^{-2t}$
Initial voltage is all across the inductor.
    $V_{L}(0)=1=A$
    $A=1$
At this point will have to do integral .. to get to the current. There is no other way to use the known initial conditions: current (initially zero), and V$_C$ (initially zero). Will have to introduce integration constant and then evaluate that. More chance of mistakes, more complex, so start over with something else.
    $i(t)={\frac {1}{L}}\int _{0}^{t}V_{L}(t)dt=\int _{0}^{t}(e^{-2x}-B(x)e^{-2x})dx$
    $i(t)=B*({\frac {e^{-2t}(2t+1)}{4}}-{\frac {1}{4}})-{\frac {e^{-2t}}{2}}+{\frac {1}{2}}+C_{1}$
    $i(0)=0=C_{1}$
Ok so C$_1$ is zero. Now need to find B. Find B by doing another integral to get V$_C$:
    $V_{C}(t)={\frac {1}{C}}\int _{0}^{t}i(t)dt=4*\int _{0}^{t}(B*({\frac {e^{-2t}(2t+1)}{4}}-{\frac {1}{4}})-{\frac {e^{-2t}}{2}}+{\frac {1}{2}})dx$
    $V_{C}(t)=B+2t+e^{-2t}-Bt-Be^{-2t}-Bte^{-2t}-1+C_{1}$
    $V_{C}(0)=0=B+1-B-1+C_{1}$
    $C_{1}=0$
Still doesn't help us find B. Guess B = 2 since (2t-Bt) has to equal zero if V$_C$ is going to converge on 1. Then see if V$_C$(∞) = 1:
    $V_{C}(t)=2+e^{-2t}-2e^{-2t}-2te^{-2t}-1$
At t = ∞ what is the 2te$^{-2t}$ term's value?
Mupad says 0.
    $V_{C}(\infty )=2-1=1$
Yes! B = 2 works ... looks like the only thing that works .... So:
    $i(t)=2*({\frac {e^{-2t}(2t+1)}{4}}-{\frac {1}{4}})-{\frac {e^{-2t}}{2}}+{\frac {1}{2}}=te^{-2t}$
This means that V$_R$ is:
    $V_{R}(t)=4*i(t)=4te^{-2t}$

### Impulse Solution

Taking the derivative of the above get:
    $V_{R}\delta (t)=4e^{-2t}-8te^{-2t}$

### Convolution Integral

    $V_{R}(t)=\int _{0}^{t}(4e^{-2(t-x)}-8(t-x)e^{-2(t-x)})(2x-3x^{2})dx$
    $V_{R}(t)=8-8e^{-2t}-10te^{-2t}-6t$
There will not be any constant since again, V_R(t) = 0 after a long time ... and the capacitor opens.
# Circuit Theory/Convolution Integral/Examples/Example43

Given that i$_s$ = 1 + cos(t), find i$_o$ using the convolution integral.
Outline:

### Transfer Function

    $H(s)={\frac {\mathbb {I} _{o}}{\mathbb {I} _{s}}}={\frac {\mathbb {V} _{T}}{\mathbb {I} _{s}}}*{\frac {\mathbb {I} _{o}}{\mathbb {V} _{T}}}={\frac {1}{{\frac {1}{1/s}}+{\frac {1}{1}}+{\frac {1}{s+1}}}}*{\frac {1}{s+1}}$
    $H(s)={\frac {1}{s^{2}+2s+2}}$

### Homogeneous Solution

Setting the denominator to zero and find the values of s:
    $s=-1\pm i$
This means the solution has the form:
    $i_{o_{h}}(t)=e^{-t}(A\cos t+B\sin t)+C_{1}$

### Particular Solution

After a long time the cap opens and the inductor shorts. The current is split between the two resistors. Each will get 1/2 of the unit step function source which would be 1/2 amp:
    $i_{o_{p}}(t)={\frac {1}{2}}$

### Initial Conditions

    $i_{o}={\frac {1}{2}}+e^{-t}(A\cos t+B\sin t)+C_{1}$
The particular solution still has to apply so at t= ∞:
    $i_{o}(\infty )={\frac {1}{2}}={\frac {1}{2}}+C_{1}$
    $C_{1}=0$
Initially the current has to be zero in this leg so:
    $i_{o}(0)={\frac {1}{2}}+A=0$
    $A=-{\frac {1}{2}}$
The initial voltage across the cap is zero, and across the leg is zero and across the inductor is zero. So:
    $V_{L}(t)=L*{di_{o}(t) \over dt}$
    $V_{L}(t)=e^{-t}(B(cos(t)-sin(t))+{\frac {1}{2}}(cos(t)+sin(t))$
    $V_{L}(0)=B+{\frac {1}{2}}=0$
    $B=-{\frac {1}{2}}$
So now:
    $i_{o}={\frac {1}{2}}(1-e^{-t}(\cos t+\sin t))$

### Impulse Response

Taking the derivative of the above
get:
    $i_{o_{\delta }}(t)=e^{-t}\sin t$

### Convolution Integral

    $i_{o}(t)=\int _{0}^{t}e^{-t+x}\sin(t-x)*(1+cos(x))dx$
    $i_{o}(t)=0.2\cos(t)+0.4\sin(t)-0.7\cos(t)e^{-t}-1.1\sin(t)e^{-1}+0.5$
# Circuit Theory/Convolution Integral/Examples/Example42

Given that V$_s$ = (1 + 2t)μ(t), find i$_o$ using the convolution integral.

### Transfer function

    $H(s)={\frac {\mathbb {I} _{o}}{\mathbb {V} _{s}}}={\frac {\mathbb {I} _{T}}{\mathbb {V} _{s}}}*{\frac {\mathbb {I} _{o}}{\mathbb {I} _{T}}}={\frac {1}{s+{\frac {1}{1+{\frac {1}{1+s}}}}}}*{\frac {\frac {1}{1+s}}{1+{\frac {1}{1+s}}}}$
    $H(s)={\frac {1}{s^{2}+3s+1}}$

### Homogeneous Solution

Two real, different roots:
    $s_{1,2}={\frac {-3\pm {\sqrt {5}}}{2}}$
So I$_o$ has this form:
    $i_{o}(t)=Ae^{s_{1}t}+Be^{s_{2}t}$

### Particular Solution

After a very long time, inductors are shorts, the voltage across both 1 ohm resistors is 1 volt, so i$_o$ is 1 amp:
    $i_{o_{p}}(t)=1$

### Initial Conditions

Adding the particular and homogeneous solutions together, have this form:
    $i_{o}(t)=1+Ae^{s_{1}t}+Be^{s_{2}t}+C_{1}$
After a long period of time (using the particular initial condition again), the current is going to be 1:
    $i_{o}(\infty )=1+C_{1}=1$
    $C_{1}=0$
Initially, the two inductors are going to be opens, thus i$_o$ has to be 0:
    $i_{o}(0)=0=1+A+B$
Initially, all the drop is going to be across the first inductor, leaving the voltage across the second zero. The current i$_o$ flows through both the second inductor and it's serial resistor, so an equation for this voltage can be obtained by taking the derivative:
    $V_{L}(t)=L{di_{o}(t) \over dt}=s_{1}Ae^{s_{1}t}+s_{2}Be^{s_{2}t}$
    $V_{L}(0)=0=s_{1}A+s_{2}B$
    $A={\frac {s_{2}}{s_{1}-s_{2}}}$
    $B=-{\frac {s_{1}}{s_{1}-s_{2}}}$
So:
    $i_{0}(t)=1+{\frac {s_{2}e^{s_{1}t}-s_{1}e^{s_{2}t}}{s_{1}-s_{2}}}$

### Impulse Solution

Taking the derivative of the above get:
    $i_{o_{\delta }}(t)={\frac {s_{1}s_{2}}{s_{1}-s_{2}}}(e^{s_{1}t}-e^{s_{2}t})$

### Convolution Integral

    $i_{o}(t)=\int _{0}^{t}{\frac {s_{1}s_{2}}{s_{1}-s_{2}}}(e^{s_{1}t}-e^{s_{2}t})*(1+2x)dx$
    $i_{o}(t)=2t+4.96e^{-0.382t}+0.0403e^{-2.62t}-5+C_{1}$
At t=0:
    $i_{o}(0)=0=4.96+.04-5+C_{1}$
Which means that C_1 = 0 so finally:
    $i_{o}(t)=2t+4.96e^{-0.382t}+0.0403e^{-2.62t}-5$
# Circuit Theory/Convolution Integral/Examples/example49/Vc

Given that the source voltage is (2t-3t$^2$), find voltage across the resistor.
This is the V$_c$ solution.
Outline:

### Transfer Function

    $H(s)={\frac {V_{C}}{V_{S}}}={\frac {\frac {1}{0.25s}}{4+s+{\frac {1}{0.25s}}}}$
    $H(s)={\frac {4}{s^{2}+4s+4}}$

### Homogeneous Solution

There are two equal roots at s = -2, so the solution has the form:
    $V_{C_{h}}(t)=Ae^{-2t}+Bte^{-2t}+C_{1}$

### Particular Solution

After a long time attached to a unit step function source, the inductor has shorted and the capacitor has opened. All the drop is across the capacitor. The current is zero.
    $V_{C_{p}}=1$
This also means that C$_1$ is still unknown.

### Initial Conditions

So far the full equation is:
    $V_{C}(t)=1+Ae^{-2t}+Bte^{-2t}+C_{1}$
At t = ∞ what is the B term's value?
Mupad says 0. So This means that at t = ∞:
    $1=1+C_{1}$
    $C_{1}=0$
Initial voltage across the capacitor is 0 so:
    $0=1+A+C_{1}$
    $A=-1$
Initial current through the series leg is zero because of the assumed initial conditions of the inductor. This means i(0) = 0, so:
    $i(t)=C*{dV_{C}(t) \over dt}={\frac {1}{4}}((-2A+B)e^{-2t}-2Bte^{-2t})$
    $0={\frac {1}{4}}(-2A+B)$
    $B=2A$
    $B=-2$
This means that i(t) =:
    $i(t)={\frac {1}{4}}((-2(-1)+(-2))e^{-2t}-2(-2)te^{-2t})=te^{-2t}$
    $V_{R}(t)=4*i(t)=4te^{-2t}$

### Impulse Solution

Taking the derivative of the above get:
    $V_{R}\delta (t)=4e^{-2t}-8te^{-2t}$

### Convolution Integral

    $V_{R}(t)=\int _{0}^{t}(4e^{-2(t-x)}-8(t-x)e^{-2(t-x)})(2x-3x^{2})dx$
    $V_{R}(t)=8-8e^{-2t}-10te^{-2t}-6t$
There will not be any constant since again, V_R(t) = 0 after a long time ... and the capacitor opens.
# Circuit Theory/Cover

Circuit Theory
Wikibooks: The Free Library
# Circuit Theory/Convolution Integral/Examples/example49

Given that the source voltage is (2t-3t$^2$), find voltage across the resistor.
Can do focused on V$_r$ or: current, V$_c$, or V$_L$ before converting to V$_r$ .. Below is the V$_R$ solution.
Outline:

### Transfer Function

    $H(s)={\frac {V_{R}}{V_{S}}}={\frac {4}{4+s+{\frac {1}{0.25s}}}}$
    $H(s)={\frac {4s}{s^{2}+4s+4}}$

### Homogeneous Solution

There are two equal roots at s = -2, so the solution has the form:
    $V_{R_{h}}(t)=Ae^{-2t}+Bte^{-2t}+C_{1}$

### Particular Solution

After a long time attached to a unit step function source, the inductor has shorted and the capacitor has opened. All the drop is across the capacitor. The current is zero.
    $V{R_{p}}=0$
This also means that C$_1$ has to be zero.

### Initial Conditions

So far the full equation is:
    $V_{R}(t)=Ae^{-2t}+Bte^{-2t}$
Initial current through the series leg is zero because of the assumed initial conditions of the inductor. This means V$_R$ = 0:
    $V_{R}(0)=0=A$
    $A=0$
Assuming the initial voltage across the capacitor is zero, no current is flowing so the drop across the resistor is zero.
    $i(t)={\frac {V_{R}}{4}}$
    $V_{L}(t)=L{di(t) \over dt}={\frac {1}{4}}((-2A+B)e^{-2t}-2Bte^{-2t})$
    $V_{L}(0)=1={\frac {B}{4}}$
    $B=4$
    $V_{R}(t)=4te^{-2t}$

### Impulse Solution

Taking the derivative of the above get:
    $V_{R}\delta (t)=4e^{-2t}-8te^{-2t}$

### Convolution Integral

    $V_{R}(t)=\int _{0}^{t}(4e^{-2(t-x)}-8(t-x)e^{-2(t-x)})(2x-3x^{2})dx$
    $V_{R}(t)=8-8e^{-2t}-10te^{-2t}-6t$
There will not be any constant since again, V_R(t) = 0 after a long time ... and the capacitor opens.
# Circuit Theory/Convolution Integral/Examples/example49/current

Given that the source voltage is (2t-3t$^2$), find voltage across the resistor.
Here focused on finding current first:

### Transfer Function

    $H(s)={\frac {i}{V_{S}}}={\frac {1}{4+s+{\frac {1}{0.25s}}}}$
    $H(s)={\frac {s}{s^{2}+4s+4}}$

### Homogeneous Solution

There are two equal roots at s = -2, so the solution has the form:
    $i_{h}(t)=Ae^{-2t}+Bte^{-2t}+C_{1}$

### Particular Solution

After a long time attached to a unit step function source, the inductor has shorted and the capacitor has opened. All the drop is across the capacitor. The current is zero.
    $i_{p}=0$

### Initial Conditions

So far the full equation is:
    $i(t)=Ae^{-2t}+Bte^{-2t}+C_{1}$
Initial current through the series leg is zero because of the assumed initial conditions of the inductor. This means:
    $i(0)=0=A+C_{1}$
Assuming the initial voltage across the capacitor is zero, then initial voltage drop has to be across the inductor.
    $V_{L}(t)=L{di(t) \over dt}=(-2A+B)e^{-2t}-2Bte^{-2t}$
    $V_{L}(0)=1=-2A+B$
After a long period of time, the current still has to be zero so:
    $C_{1}=0$
This means that:
    $A=0$
    $B=1$
    $i(t)=te^{-2t}$
    $V_{r}(t)=4te^{-2t}$
The 4 is lost in the numerator of the transfer function if a transfer function is written for V$_r$ initially. The 4 does not make it into the homogeneous solution. In second order analysis, never write a transfer function for a resistor.

### Impulse Solution

Taking the derivative of the above get:
    $V_{R}\delta (t)=4e^{-2t}-8te^{-2t}$

### Convolution Integral

    $V_{R}(t)=\int _{0}^{t}(4e^{-2(t-x)}-8(t-x)e^{-2(t-x)})(2x-3x^{2})dx$
    $V_{R}(t)=8-8e^{-2t}-10te^{-2t}-6t$
There will not be any constant since again, V_R(t) = 0 after a long time ... and the capacitor opens.
# Circuit Theory/Convolution Integral/Examples/example48

Given that V$_s$(t) = 3μ(t) mV, and the I$_L$(0-) = 1 ma, find the voltage across the one ohm resistor.
Outline:
* find transfer function/differential equation
* find homogeneous solution
* find particular solution for step function
* evaluate initial conditions
* find impulse solution from step function solution
* use convolution integral to find general solution
* evaluate integration constant

### Transfer Function

From inspection can see the following:
    $V_{TR}=V_{R}={\frac {V_{T}}{2}}$
So:
    $H(s)={\frac {V_{tr}}{V_{s}}}={\frac {V_{tr}}{V_{s}}}={\frac {V_{T}}{V_{s}}}{\frac {V_{tr}}{V_{T}}}={\frac {1}{1+s}}*{\frac {1}{2}}$

### Homogeneous Solution

Setting the denominator equal to zero, can see that the time constant is -1. Thus the solution is of the form:
    $V_{TR_{h}}=A*e^{-t}+C_{1}$

### Particular Solution

After a long period of time hooked to a DC voltage source, an inductor appears as a short. This means that all 3mV is going to drop across the equivalent of a 1 ohm resistor. Half of this is going be V$_{TR}$. So the particular solution is going to be:
    $V_{TR_{p}}=1.5mV$

### Initial Conditions

The initial current is going to be 1 mA. This is going to be dropped by the 1 ohm resistor creating V$_T$, thus V$_T$(0+) = 1 mV and V$_{TR}$ is going to be 0.5 mV. This means that:
    $V_{TR}(t)=A*e^{-t}+C_{1}+1.5mV$
    $V_{TR}(0)=0.5mV=A+C_{1}+1.5mV$
    $A+C_{1}=-1mV$
The inductor obliges the rest of the circuit and drops the other 2mV so the source is happy. This is the source of the other initial condition. First have to turn V$_{TR}$ into an expression of total current which is the inductor current:
    $i_{TR}={\frac {V_{TR}}{1}}=V_{TR}$
Because the current is split in half has it goes through the resistors, the total current is twice that of i_{TR}:
    $i_{L}=2*V_{TR}=2*(A*e^{-t}+C_{1}+1.5m)$
From the inductor terminal relation, know that the derivative of this times inductance is going to equal the initial 2 mV across the inductor:
    $V_{L}(t)=L*{di_{L} \over dt}=L*-2Ae^{-t}$
    $V_{L}(0)=2m=L*-2A=-2*A$
    $A=-1m$
This means that C$_1$ is zero. So the answer is:
    $V_{TR}=1.5-e^{-t}mV$
# Coding Cookbook

This book features all kinds of short coding samples, often called code snippets. It is not a reference for any specific coding language, or a command reference.
Like an ordinary cookbook, you are not looking for a recipe that can be made from a certain ingredient or using a certain utensil, but you are looking to perform a specific task.
Often, these snippets are multi-disciplinary. Especially with web programming languages, you often use a combination of HTML, CSS, JavaScript and possibly server-side languages to perform a certain task. Also, the same result can often be achieved using different languages. Each task can list solutions in several languages.
By Task - By Language - Alphabetic listing
    See also other Cookbooks.
# Coding Cookbook/SQL Encode

Encode various datatypes for safe usage in SQL commands.
# Coding Cookbook/Escape from encapsulation
## JavaScript solution

Escape from frame encapsulation
# Coding Cookbook/Inline If

Not every language offers an Inline If function.
# Coding Cookbook/Change Query Parameter

Change parameters in an URL querystring (for example "param1=foo&param2=bar").
# Coding Cookbook/Initialization

Initialization is the first stage that a computer program performs to prepare itself for operation. A computer programmer may use a basic template that contains the initialization code for the program.

## Stages of initialization

The initialization stage of a program includes:
* Population of variables with default values
* Preliminary system checks
* Interpretation of command line parameters
* Consideration of the environment
* Processing of the configuration files
* Memory and resource reservation
* Service registration
* Secondary system checks
* Freeing the initialization code segment

## Procedural programming languages

For a procedural programming language, the initialization code usually follows the declaration headers, or at the beginning of the main startup function or within constructor blocks, depending on the programming language being used.
# Coding Cookbook/JavaScript Encode

Encode special characters in a string, so that it can be displayed using JavaScript
# Coding Cookbook/HTML Encode

Convert an object (commonly a string) to HTML-safe code.

## VBScript solution

Can convert strings and recordsets (converted to tables).
# Croatian/Main Contents/Level 1/Lesson 11

Following is a list of Croatian words for body parts with plural forms. The Croatian word for body is tijelo (pl. tijela).

## Humans

Nose: Nos, pl. nosevi
Ear: Uho, pl. uši (this is also plural of uš which means "lice")
Mouth: Usta (plural only {pluralia tantum})
Hair: Kosa (plural only {pluralia tantum})
Lips: Usna, pl. usne
Hips: kuk pl. kukovi, bok pl. bokovi
Hand, arm: Ruka, pl. ruke
Palm: Dlan, pl. dlanovi
Shoulder: Rame, pl. ramena
Face: lice, pl. lica
Nail: Nokat, pl. nokti
Finger: Prst, pl. prsti
Toe: Nožni prst (means : leg finger)
Scar: Ožiljak, pl. ožiljci
Bruise: Masnica or modrica (Note: derived from the word modro meaning "blue") pl. masnice, modrice
Tooth: Zub, pl. zubi
Foot: Stopalo, pl stopala

## Animals

Animal hair: Dlaka, pl. dlake
Fur, pelt: Krzno, pl. krzna
Fang: Očnjak, pl. očnjaci
Wing: Krilo, pl krila
Leg: Noga, pl noge
Muzzle: Gubica, pl. gubice
Snout: Njuška, pl. njuške
Paw: Šapa, pl. šape
Pad (on paw): Jastučić (lit. small pillow), pl. jastučići
Tail: Rep, pl. repovi (can also mean ponytail)
Whisker: Brk (also means: moustache), pl. brkovi or brci
Claw: Pandža, pl. pandže
Next Lesson
# Croatian/Main Contents/Level 1/Lesson 12

Following is a list of Croatian words for food with plural forms. Croatian word for food is hrana.

## Fruit

Fruits: Voće (plural only)
Apple: Jabuka
Pear: Kruška
Banana: Banana
Blueberry: Borovnica
Currants: Ribizla, pl. ribizle
Black currants: Crni ribizl, pl. crni ribizli
Raspberry: Malina
Apricot: Marelica
Peach: Breskva
Fig: Smokva
Grapes: Grožđe
Almond: Badem
Strawberry: Jagoda, pl. jagode
Cherry: Trešnja
Sour cherry: Višnja
Blackberry: Kupina
Pomegranate: Šipak
Pineapple: Ananas
Melon: Dinja
Quince: Dunja
Raisin: Grožđica
Watermelon: Lubenica
Kiwi: Kivi
Grapefruit: Grejp
Orange: Naranča

## Vegetables

Vegetable: Povrće (pl. only)
Potato: Krumpir
Tomato: Rajčica
Aubergine, eggplant: Patlidžan
Courgettes, zucchini: Tikvice
Cucumber: Krastavac, pl. Krastavci
Mushrooms: Gljive, šampinjoni
Cauliflower: Karfiol, cvjetača
Broccoli: Brokula
Chard: Blitva
Pumpkin: Bundeva, buča
Celery: Celer
Garlic: Češnjak, bijeli luk
Brussels sprouts: Prokulica
Kale, savoy: Kelj
Beet, turnip: Repa
Kohlrabi: Koraba
Beans: Grah (it has only a singular form)
Green beans: Mahune (a singular form is mahuna, but it is almost never used)
Paprika: Paprika
Carrot: Mrkva
Onion: Luk
Leek: Poriluk
Peanuts: Kikiriki
Spinach: Špinat
Pea: Grašak
Corn, maize: Kukuruz
Hazelnut: Lješnjak
Walnut: Orah
Olive: Maslina
Parsley: Peršin
Radish: Rotkva
Lettuce: Salata

## Meals and Drinks

Cheese: Sir
Meat: Meso
Soup: Juha
Bacon: Šunka
Butter: Maslac
Eggs: Jaja
Beer: Pivo
Bread: Kruh
Coffee: Kava
Juice: Sok
Tea: Čaj
Milk: Mlijeko
Fish: Riba

## Meat

Beef: Govedina
Venison: Divljač, srnetina
Veal: Teletina

## Fish and shelves

Fish: Riba
Shelve: Školjka
Trout: Pastrva
Mackerel: Skuša
Anchovy: Inćun
Mussels: Dagnje
Oysters: Kamenice
Sole: List
Cod-fish: Bakalar
Tuna: Tuna, tunj
Sardines: Sardina, srdela
Lobster: Jastog
Salmon: Losos
Prawn: Kozica, rak
Shrimp: Morski račić
Next Lesson
# Croatian/Main Contents/Level 1/Lesson 14

Nouns are never used instead (For example: Fox fur).
Sweet: Sladak
Hot (temperature): vruć
Spicy: Ljut
Cold: Hladan
Boring: Dosadan
Angry: Ljut
Cute, sweet: Sladak
Ugly: Ružan
Beautiful: Lijep
Scary: Strašan
Simple: Jednostavan
Hard: Težak
Light: Lagan
Heavy: Težak
Dangerous: Opasan
Strict: Strog
Round: Okrugao
Fat: Debeo
Skinny: Mršav
New: Nov
Old: Star
Dead: Mrtav
Alive: Živ
Next Lesson
# Croatian/Main Contents/Level 1/Lesson 10

Money : Novac
Bank-note : Novčanica
Coin : Kovanica (kovati means forging)
Croatian money
    Kuna (0,14 of Euro, lit. marten)
    Lipa (100 Lipa = 1 kuna lit. linden)
Next Lesson
# Croatian/Main Contents/Level 1/Lesson 13

Willow : Vrba
Oak tree: Hrast (acorn : žir)
Beech : Bukva
Pine : Bor
Fir tree : Jela
Ash : Jasen
Maple : Javor
Next Lesson
# Croatian/Main Contents/Level 1

These lessons are not mandatory, as it would take a very long time to learn all the words in these lessons. These lessons are more of a dictionary, and are to be used for general help. However, it is recommended to learn Lesson 1, Lesson 2, and Lesson 3, as these lessons will help for the lessons in level 2.
Lesson 1 - The Alphabet and stress/Abeceda i naglasak
Lesson 2 - Common Words/Uobičajene riječi
Lesson 3 - Question Words/Upitne riječi
Lesson 4 - Animals/Životinje
Lesson 5 - Months/Mjeseci
Lesson 6 - Colours/Boje
Lesson 7 - School/Škola
Lesson 8 - Family/Obitelj
Lesson 9 - Seasons/Sezone
Lesson 10 - Money/Novac
Lesson 11 - Body/Tijelo
Lesson 12 - Food/Hrana
Lesson 13 - Plants/Biljke
Lesson 14 - Adjectives/Pridjevi
Lesson 15 - Verbs/Glagoli
Lesson 16 - Expressions, Phrases/Izrazi, fraze
Lesson 17 - Prepositions/Prijedlozi
Lesson 18 - Numbers/Brojevi
Lesson 19 - Nationalities and ethnicities in Croatian/Nacionalnosti i etničke pripadnosti
# Civic Technology

This book introduces the reader to common uses of civic technology, and ways ordinary citizens can improve the digital wellbeing of their local community.
This book gives the reader a number of example exercises they can use to help their community on the internet. To follow the exercises outlined within this book, all that is needed is simple computer knowledge. No exercise in this book require coding knowledge, or hardware more expensive than a basic computer or basic smartphone.

## Table of Contents

This book contains a collection of primers with activities you can pursue to help your community digitally.
* Give Directions
* Preserve History
* Speak your language
* Teach

## Further Reading

* FreedomBox for Communities
# Civ/Sid Meier's Alpha Centauri/Understanding the Governor
## The Governor

The Governor is a means by which players can automate or partially automate the operation of their cities. The alternative is to manually control every aspect of it.

### Default Setting

By default all the options are enabled but the governor is not active. Clicking on any of the Build, Conquer, Discover or Explore options in the city control screen will activate the Governor. When the governor is activated in this manner the game engine is in complete control of the city and it will follow the priorities set by the Build, Conquer, Discover or Explore options.

### Running with Minimal Options

More advanced players may wish to run with a more minimal set of options. For example, running with just the "Governor is active" and "Governor manages citizens and specialists (including Drone Riots)" will leave the governor managing the citizens - including assigning workers to tiles and dealing with drone riots - while leaving the human player in control of the build queue. Running with these settings also exerts some control over which tiles are assigned to; for example selecting Explore or Discover will result in workers primarily being assigned to tiles producing food; selecting Build will result in more forest squares (with their additional minerals) being included.

### Running without the Governor

At higher levels the game engine has a big advantage over human players. To some extent this can be partially negated by micro-managing cities in their early stages. For example, by manually assigning workers it is possible to ensure that the city builds a Recreation commons just as it increases from size 2 to size 3. The Recreation Commons are neither built early (which wastes energy) not late (which can result in drone riots). However, micro-managing the first cities is labour intensive and may detract from enjoyment of the game.